Learning Objectives

After going through this, unit you will be able to understand, appreciate and apply the following concept:

• Identify different type of charges.
• Determine the electric force between two point charges using Coulomb’s Law.
• Apply Coulambl’s law to a system of point charges to determine the net force on a charge.
• Calculate the electric field due to a point charge or system of point charges.
• Determine the electric flux through a surface.
• Use Guass law to find electric field of various charge distribution.
• Determine the torque experienced by an electric dipole in a uniform electric field.
• Distinguish between electric potential and electric potential energy.
• Calculate the electric potential of a configuration of point charges.
• Determine how electric field strength is related to electric potential and electrostatic potential energy.
• Calculate the capacitance of a given electric network, energy stored in a parallel plate capacitor with and without dielectric.

Concept Map of Electrostatics

$~$

Electrostatics

Electrostatics is the branch of physics that deals with the phenomena and properties associated with charges at rest.

Electric Charge

Electric charge is an intrinsic characterstics of the fundamental particles of matter giving rise to all electric and magnetic forces and interactions.

From simple experiments on frictional electricity, it has been known that some materials, such as amber or ebonite, on being rubbed with, wool get ’electrically charged’. Similarly, when a glass rod is rubbed with silk cloth, the glass rod gets ’ electrically charged’. However, the charges of these two rods are of different types. One can infer that there are two types, of charges in nature: ‘glass rod type’ and ’ebonite rod type’. The first type has been called as ‘positive charge’ while the second type has been called the ’negative charge’. The changes were named as ‘positive’ and ’negative’ by the American scientist, Benjamin Fraklin, in 1750. When the ‘positive’ and ’negative’ charges on an object, are not in equal amounts, there is a net charge on the object, that can interact with other objects. We regard such an object as an ’electrified’ or " ‘charged’ object. When an object has no net charge, it is said to be electrically neutral.

It has been observed that: “Like charges repel each other and unlike charges attract each other.”

Basic Properties of Electric Charge

1. Charge is quantised : Electric charge can have only discrete values. The smallest discrete value that can exist in nature is the charge on the electron, ’ $e$ ‘. Charge on any body always exists in integral multiples of this fundamental ‘unit’ of electric charge.

$$ \mathrm{q}= \pm \mathrm{n} \text { e where } \mathrm{n}= \pm 1, \pm 2 \ldots $$

2. Charge is conserved : For an isolated system, its total charge remains constant. “charge is neither created nor distroyed”, it can only be tansferred from one object to the other.

3. Charge is additive: If a system contains ’ $n$ ’ charges; $q _{1}, q _{2}, q _{3} \ldots . q _{n}$; the total charge of the sytsem is:

$$ \mathrm{q} _{\mathrm{net}}=\left(\mathrm{q} _{1}+\mathrm{q} _{2}+\mathrm{q} _{3}+\ldots+\mathrm{q} _{\mathrm{n}}\right) $$

Charges thus add up like ‘real numbers’, they are scalars like mass.

4. Charge is invariant : The numerical value of the charge on an isolated object is independent of its state of rest or motion.

The S.I. Unit of charge is the ‘Coulomb’. It is defined in terms of ’electric current’. ‘One Coulomb’ equals the charge, transferred in one second, across the cross section of a wire carrying a current of one ampere.

Coulomb’s Law

We have noted that two like charges repel each other while two unlike charges attract each other. This force of interaction, between two charges is known as ’electric force’ or ‘coulomb force’. On the basis of experiments, coulomb gave a law for finding the magnitude of the force acting between two point charges.

According to Coulomb’s law; two stationary point charges, $\mathrm{q} _{1}$ and $\mathrm{q} _{2}$, repel, attract, each other with a force (’ $\mathrm{F}$ ‘) which is directly proportional to the product $\left(\mathrm{q} _{1} \mathrm{q} _{2}\right)$ of the two of charges and inversely propotional to the square of the distance (’ $r$ ‘) between them

$$ \therefore \mathrm{F} \alpha\left(\frac{\mathrm{q} _{1} \mathrm{q} _{2}}{\mathrm{r}}\right) $$

or $\quad \mathrm{F}=\mathrm{K} \frac{\mathrm{q} _{1} \mathrm{q} _{2}}{\mathrm{r}^{2}}$ where $\mathrm{K}$ is a proportionality constant; its value depends upon the nature of the medium separating the two charges. When the charges are placed in vacuum (or air) we take this proportionality constant as $\frac{1}{4 \pi \varepsilon _{0}}$. Thus $\mathrm{F}=\frac{1}{4 \pi \varepsilon _{0}} \frac{\mathrm{q} _{1} \mathrm{q} _{2}}{\mathrm{r}^{2}}$.

The constant $\varepsilon _{0}$ is called permittivity of free space or vacuum. In SI Units, its value is

$$ \begin{aligned} & \mathrm{K}\left(=\frac{1}{4 \pi \varepsilon _{0}}\right)=9 \times 10^{9} \mathrm{Nm}^{2} / \mathrm{C}^{2} \\ & \therefore \varepsilon _{0} \simeq 8.85 \times 10^{-12} \mathrm{C}^{2} \mathrm{~N}^{-1} \mathrm{~m}^{-2} \end{aligned} $$

For mediums other than vacuum (or air) the force between the charges is given by:

$$ \mathrm{F}=\frac{1}{4 \pi \varepsilon _{\mathrm{m}}} \frac{\mathrm{q} _{1} \mathrm{q} _{2}}{\mathrm{r}^{2}} $$

Here $\varepsilon _{\mathrm{m}}$ is called the permittivity of the medium concerned. For materials other than vaccum(or air), we can put

$$ \varepsilon _{\mathrm{m}}=\mathrm{k} \varepsilon _{0} $$

Here $\mathrm{k}$ is a dimensionless constant. It is called as the ‘dielectric constant’ or ‘relative permittivity’, of the medium.

The above expression can, therefore, be written as

$$ \mathrm{F}=\frac{1}{\mathrm{k}}\left(\frac{1}{4 \pi \varepsilon _{0}}\right) \frac{\mathrm{q} _{1} \mathrm{q} _{2}}{\mathrm{r}^{2}} $$

Important Points Regarding Coulomb’s Law

1. The Coulomb force, acts along the straight line, connecting the ponits of location of the two ‘point’ charges.

2. The law can be used to define the SI unit of charge the ‘Coulomb’.

3. For a point charge, the coulomb force due to it is ‘central’ and has a spherical symmetry.

4. For materials media the value of ’ $k$ ’ is greater than 1 . Thus, if two charges kept a given distance apart, have a medium other than vacuum (or air), between them the force, between them, decreases by a factor $\mathrm{k}$.

Coulomb’s Law in Vector Form

$~$

It $\mathrm{q} _{1}$ and $\mathrm{q} _{2}$ be two point charges separated by a distance $\mathrm{r}$ in vacuum, the force on $\mathrm{q} _{2}$ due to $\mathrm{q} _{1}\left(\mathbf{F} _{21}\right)$ is given by:

$$ \mathbf{F} _{21}=\frac{1}{4 \pi \varepsilon _{0}} \frac{\mathrm{q} _{1} \mathrm{q} _{2}}{\mathrm{r}^{2}} \hat{\mathrm{r}} _{12} $$

Where $\hat{r} _{12}$ is a unit vector, pointing from $\mathrm{q} _{1}$ to $\mathrm{q} _{2}$.

The force, on $\mathrm{q} _{1}$ due to $\mathrm{q} _{2}$, is given by

$$ \mathbf{F} _{21}=\frac{1}{4 \pi \varepsilon _{0}} \frac{\mathrm{q} _{1} \mathrm{q} _{2}}{\mathrm{r}^{2}} \hat{\mathrm{r}} _{21} $$

Here $\hat{\mathrm{r}} _{21}$ is a unit vector along the direction of the line directed from $\mathrm{q} _{2}$ to $\mathrm{q} _{1}$

$\hat{\mathbf{r}} _{12}$ and $\hat{\mathrm{r}} _{21}$ are clearly unit vectors in opposite directions. We can therefore, write

$$ \mathbf{F} _{21}=-\mathbf{F} _{21} $$

In general, one can write:

$$ \mathbf{F}=\frac{1}{4 \pi \varepsilon _{0}} \frac{\mathrm{q} _{1} \mathrm{q} _{2}}{\mathrm{r}^{2}} \hat{\mathrm{r}} $$

$$ \begin{aligned} & =\frac{1}{4 \pi \varepsilon _{0}} \frac{\mathrm{q} _{1} \mathrm{q} _{2}}{\mathrm{r}^{3}} \hat{\mathrm{r}} \\ & =\frac{1}{4 \pi \varepsilon _{0}} \frac{\mathrm{q} _{1} \mathrm{q} _{2}}{\mathrm{r}^{3}}(x \hat{\mathrm{i}}+\mathrm{y} \hat{\mathrm{j}}+\mathrm{z} \hat{\mathrm{k}}) \end{aligned} $$

Here $\hat{r}$ is the unit vector that would have ‘its origin’ at the source of the force. A positive magnitude of $|\mathbf{F}|$, implies repulsion while a negative value of $|\mathbf{F}|$, implies attraction between the two charges.

How to Solve Problems using Coulomb’s Law

1. Find the magnitude of force, using Coulomb’s law, ignoring the signs of charges. Signs only indicate the direction of force.

2. Decide whether the force, due to a given charge, is attractive or repulsive, express it by drawing vectors.

Principle of Superposition of Charges

When more than two charges are present in space, Coulomb’s law holds for every pair of charges. Thus, the net force on any charge is the ‘vector’ sum of the forces exerted on it by all other charges, taking one charge at a time.

This is known as the principle of super position for charges.

Thus if there are $(\mathrm{n}+1)$ point charges, the force, one of them, say $\mathrm{q} _{1}$ due to all other charges is given by.

$~$

$$ \mathbf{F} _{1}=\mathbf{F} _{12}+\mathbf{F} _{13}+\ldots+\mathbf{F} _{n} $$

Here the ‘individual forces’ are unaffected due to the presence of other charges.

or $\therefore \mathbf{F} _{1}=\left(\frac{1}{4 \pi \varepsilon _{0}} \mathrm{q} _{1}\right) \sum _{\mathrm{i}=2}^{\mathrm{n}} \frac{\mathrm{q} _{\mathrm{i}}}{\mathrm{r} _{\mathrm{i} 1}^{2}} \mathbf{r} _{\mathrm{i} 1}$

Here $\vec{r} _{i 1}$ is the unit vector along the direction joining the charge $q _{i}$ to the charge $q _{1}$.

Example-(1):

A given charge $\mathrm{Q}$ is to be shared between two (fixed) objects kept at given distance apart. Find the values of the charges, on the two objects, so that the force between them has its maximum values.

Show Answer

Solution:

Let charge on one object be ’ $q$ ‘; The charge on other, will then be Q-q

$$ \therefore \mathrm{F}=\frac{1}{4 \pi \varepsilon _{0}} \frac{\mathrm{q}(\mathrm{Q}-\mathrm{q})}{\mathrm{r}^{2}} $$

$~$

For $\mathrm{F}$ to be maximum, we need to have

$$ \begin{gathered} \frac{\mathrm{dF}}{\mathrm{dq}}=0 \\ \therefore \frac{\mathrm{d}}{\mathrm{dq}}\left(\frac{1}{4 \pi \varepsilon _{0}} \frac{(\mathrm{qQ})}{\mathrm{r}^{2}}-\frac{1}{4 \pi \varepsilon _{0}} \frac{\left(\mathrm{q}^{2}\right)}{\mathrm{r}^{2}}\right)=0 \\ =\frac{\mathrm{Q}}{4 \pi \varepsilon _{0} \mathrm{r}^{2}}-\frac{1}{4 \pi \varepsilon _{0}} \frac{2 \mathrm{q}}{\mathrm{r}^{2}}=0 \\ =\mathrm{q}=\frac{\mathrm{Q}}{2} \end{gathered} $$

Thus, the charge should equally divided between the two objects.

Example-(2):

Two ‘free’ point charges, $(+4 \mathrm{e})$ and (+e), are placed a distance ’ $\mathrm{a}$ ’ apart. Where should a third charge, (say q), be placed, between them, such that the entire system may be in equilibrium? What should be, the magnitude and sign of q? State the ’type of equilibrium’ of the system.

Show Answer

Solution:

Let charges be placed as shown in the figure.

$~$

The charge $+\mathrm{e}$ exerts a repulsive force on charge $+4 \mathrm{e}$. Hence for the equilibrium of charge $+4 \mathrm{e}$ the third charge ’ $q$ ’ must exert an attractive force $F$ ’ on $+4 e$. This requires cahrge ’ $q$ ’ to be a negative charge. For equilibrium of charge (+4e).

$$ \begin{aligned} & \mathbf{F}=-\mathbf{F}^{\prime} \\ & =|\mathbf{F}|=|-\mathbf{F}| \\ & =\frac{1}{4 \pi \varepsilon _{0}} \frac{(4 \mathrm{e} \times \mathrm{e})}{\mathrm{a}^{2}}=\frac{1}{4 \pi \varepsilon _{0}} \frac{(4 \mathrm{eq})}{x^{2}} \\ & \therefore \mathrm{q}=\frac{\mathrm{ex}^{2}}{\mathrm{a}^{2}} \end{aligned} $$

For equilibrium of charge ( ‘-q )

$$ \begin{aligned} & \mathbf{F} _{1}=-\mathbf{F} _{2} \\ & \left|\mathbf{F} _{1}\right|=\left|-\mathbf{F} _{2}\right| \\ & \Rightarrow \frac{1}{4 \pi \varepsilon _{0}}\left(\frac{4 \mathrm{eq}}{x^{2}}\right)=\frac{1}{4 \pi \varepsilon _{0}} \frac{\mathrm{eq}}{(\mathrm{a}-x)^{2}} \\ & \Rightarrow x=\frac{2 \mathrm{a}}{3} \\ & \mathrm{Thus} \quad|\mathrm{q}|=\frac{\mathrm{e} x^{2}}{\mathrm{a}^{2}}=\frac{4 \mathrm{e}}{9} \\ & \therefore \mathrm{q}=\left(-\frac{4 \mathrm{e}}{9}\right) \end{aligned} $$

This charge has to be kept at a distance $\left(\frac{2 a}{3}\right)$ from the charge $(+4 e)$.

It is easy to see that the equilibrium of the system is an unstable equilibrium.

Example-(3):

Two identical spheres, having unequal opposite charges, are placed at a distance $0.9 \mathrm{~m}$ apart from each other. After touching them mutually, they are again put at the same distance apart. They are now observed to repel each other with a force of $0.025 \mathrm{~N}$. Calculate the final charge on each of the two spheres.

Show Answer

Solution:

Since the spheres are identical after touching them they will acquire similar and equal charges. Suppose the charge on each is ’ $\mathrm{q}$ ’ then by Coulomb’s law;

$$ \begin{aligned} & \mathrm{F}=\frac{1}{4 \pi \varepsilon _{0}} \frac{\mathrm{q}^{2}}{\mathrm{r}^{2}} \\ & \therefore 0.025=9 \times 10^{9} \times \frac{\mathrm{q}^{2}}{(0.9)^{2}} \\ & \mathrm{q}^{2}=2.25 \times 10^{-12} \mathrm{C}^{2} \\ & \text { or } \mathrm{q}=\sqrt{2.25 \times 10^{-12} \mathrm{C}^{2}} \\ & \text { or } \mathrm{q}= \pm 1.5 \times 10^{-6} \mathrm{C}= \pm 1.5 \mu \mathrm{C} \end{aligned} $$

Concept of Electric Field

Let us consider an electric change $q$ located in space. It we now bring another change $\left(\Delta q _{0}\right)$, near this charge $\mathrm{q}$, the charge $\left(\Delta \mathrm{q} _{0}\right)$ expriences a force of attraction or repulsion. The space, around an electric charge, $q$, in which another charge, $\left(\Delta q _{0}\right)$ experiences its (electrostatic) force of interaction, may be thought of as the electric field of charge $q$. The charge ’ $q$ ’ is known as the ‘source’ charge and charge $\left(\Delta \mathrm{q} _{0}\right)$, used to ‘detect’ the electric field (of charge $\mathrm{q}$ ) is known as the ’test’ charge.

To measure the strength or intensity of an electric field of any point, we may place a small positive charge $\left(\Delta \mathrm{q} _{0}\right)$ at that point and measure the electrostatic force (due to the source charge $\mathrm{q}^{\prime}$ ) experienced by this charge $\Delta q _{0}$. This force, per unit charge, is taken as a measure of the strength of the electric field, at that point.

That, if $(\Delta \mathbf{F})$ be the fore acting, on a test charge $\left(\Delta \mathrm{q} _{0}\right)$, placed at a given point in an elctric field, the electric field intensity, or strength, $\mathbf{E}$, of the field at that point in given by.

$$ \mathbf{E}=\lim _{\left(\Delta \mathrm{q} _{0}-0\right)}\left(\frac{\Delta \mathbf{F}}{\Delta \mathrm{q} _{0}}\right) $$

The test charge must be negligibly small so that the source charge, $q$, is not disturbed from its position and its electric field at a given point, remains unchanged. The SI unit of $\mathbf{E}$ is, therefore, $\frac{\mathrm{N}}{\mathrm{C}}$.

The electric field, due to a single point charge, kept in vacuum (or air) is given by

$$ \begin{aligned} & \mathbf{E}=\frac{1}{4 \pi \varepsilon _{0}} \frac{\mathrm{q}}{\mathrm{r}^{2}} \hat{\mathrm{r}} \\ & \therefore|\mathbf{E}|=\frac{1}{4 \pi \varepsilon _{0}} \frac{\mathrm{q}}{\mathrm{r}^{2}} \end{aligned} $$

Important Points Regarding the Electric Field

1. Once the field strenght $(\mathbf{E})$ is known, the force, on any charge ’ $q$ ‘, can be obtained from

$$ \begin{aligned} & \mathbf{F}=\mathrm{q} \mathbf{E} \\ & \therefore|\mathbf{F}|=\mathrm{q}|\mathbf{E}| \text { or } \mathrm{F}=\mathrm{qE} \end{aligned} $$

A positive charge would experience a force in the direction of the electric field while a negative charge would do so in a direction opposite to that of field.

2. Electric field, due to a system of charges, at a point, is the vector sum of the electric fields at that point due to the individual charges.

$\mathbf{E} _{\text {net }}=\mathbf{E} _{1}+\mathbf{E} _{2}+\ldots \mathbf{E} _{\mathrm{n}}$

3. Electric field, due to a point charge, points away from the charge if it is a positive charge and towards it if it is negative charge.

4. Electric field, of a point charge, has a spherical summetry.

Example-(4):

The figure shows system of two equal (in magnitude) point charges. Charge $+q$ is at $(0,1)$ and charge $-q$ is at $(1,0)$. Find the resultant field intensity at a point $\mathrm{P}$ having coordinates $(4,4)$.

$~$

Show Answer

Solution:

We have,

$$ \begin{aligned} & \mathbf{E} _{1}=\frac{1}{4 \pi \varepsilon _{0}} \frac{\mathrm{q}(4 \hat{\mathrm{i}}+3 \hat{\mathrm{j}})}{(5)^{3}} \text { units } \\ & \mathbf{E} _{2}=\frac{1}{4 \pi \varepsilon _{0}} \frac{-\mathrm{q}(4 \hat{\mathrm{i}}+3 \hat{\mathrm{j}})}{(5)^{3}} \text { units } \\ & \mathbf{E}=\mathbf{E} _{1}+\mathbf{E} _{2}=\frac{1}{4 \pi 6 _{0}} \frac{(\mathrm{q})}{25 \sqrt{5}}(\hat{\mathrm{i}}-\hat{\mathrm{j}}) \text { units } \end{aligned} $$

Continuous Charge Distributions

A system of closely spaced charges may be thought of as forming a continuous charge distribution. There are three kinds of continuous charge distributions- linear charge distribution, surface charge distribution and volume charge distribution. The electric field due to a continuous charge distribution, can be obtained in the same way as for a system of discrete charges using Coulomb’s law and the principle of super position. We first find the electric field due to small element ’ $\mathrm{dq}$ ’ and then intergrate the expression over the whole continuous charge distribution.

Example-(5):

Axial electric field due to a circular loop of charge : (charged ring)

$~$

Let us consider a uniformly charged ring of radius ’ $a$ ’ having a linear charge density ’ $\lambda$ ‘. The total charge ’ $q$ ‘, on the ring, then equals $2 \pi a \lambda$.

Show Answer

Solution:

Consider a small (line) element ’ $\mathrm{d} \ell$ ‘. It carries a charge $\mathrm{dq}=\frac{\mathrm{q}}{2 \pi \mathrm{a}} \mathrm{d} \ell$.

The magnitude of, $\mathbf{E}$, at an axial pont $\mathrm{P}$, due todq, is $\mathrm{dE}=\frac{1}{4 \pi 6 _{0}} \frac{\mathrm{dq}}{\mathrm{r}^{2}}$

Now consider another element $\mathrm{d} \ell$; diagonally opposite to $\mathrm{d} \ell$; we can see that vertical components of $\mathrm{d} \mathbf{E}$ counter balance each other in pairs. Hence the net electric field, is only due to the horizontal components of $(\mathrm{d} \mathbf{E})$ ’s.

$$ \begin{aligned} & \therefore \mathbf{E}=(\hat{\mathrm{i}}) \int \mathrm{dE} \cos \theta \\ & \therefore \mathbf{E}=\frac{1}{4 \pi \varepsilon _{0}} \frac{\mathrm{q} x \hat{\mathrm{i}}}{2 \pi \mathrm{a}\left(\mathrm{a}^{2}+x^{2}\right)^{3 / 2}} \end{aligned} $$

$$ \mathrm{Now,} \cos \theta=\frac{x}{\mathrm{r}} ~~ \text{and} ~~ \mathrm{r}=\frac{\mathrm{r}}{\sqrt{\mathrm{a}^{2}+x^{2}}} $$

$$ \mathbf{E}=\left[\frac{1}{4 \pi \varepsilon _{0}} \frac{\mathrm{q} x}{\left(\mathrm{a}^{2}+x^{2}\right)^{3 / 2}}\right] \hat{\mathrm{i}} $$

This $\overrightarrow{\mathrm{E}}$ is directed along the axial line OP.

If ’ $\mathrm{P}$ ’ lies at centre $|\mathbf{E}|=0$

If $\mathrm{P}$ is a far off axial point, i.e. if $x > > a$,

$$ |\mathbf{E}| \simeq \frac{1}{4 \pi \varepsilon _{0}} \frac{\mathrm{q}}{x^{2}} $$

Thus, for points on the axis, at distances much larger than the radius of the ring, the ring bahaves like a point charge.

Some Important Results for $\mathbf{E}$ :

• Super position of electric field - electric field at a point due to a system of charges is the vector sum of the electric fields at that point due to individual charges.

$$ \mathbf{E}=\mathbf{E} _{1}+\mathbf{E} _{2}+\ldots+\mathbf{E} _{n} $$

• The electric field due to continuous charge distribution can be obtained in much the same way as for a system of discrete charges using Coulomb’s law and the super position principle. First we find the electric field due to small element dq and then integrate the expression for whole body.

Electric field due to various (discrete / continous) charge distribution

We write below some important results for the electric field due to some charge configurations.

(i) Electric field strength due to a point charge at any point ’ $p$ ‘.

$~$

$$ \mathrm{E}=\frac{1}{4 \pi \varepsilon _{0}} \frac{\mathrm{q}}{\mathrm{r}^{2}} $$

(ii) Electric field strength due to a finite line charge with charge density ’ $\lambda$ ’ at a general ’ $\mathrm{P}$ ‘.

$~$

$$ \mathrm{E}=\frac{1}{4 \pi \varepsilon _{0}} \frac{\lambda}{x}(\sin \alpha+\sin \beta) $$

where $\lambda=\frac{\mathrm{q}}{\mathrm{L}}$, Total charge on line $=\mathrm{q}$, length $\mathrm{L}$

(iii) Electric field strength due to a infinite charge with charge density ’ $\lambda$ ’ at any general point ’ $\mathrm{P}$ ‘.

$$ \mathrm{E}=\frac{1}{4 \pi \varepsilon _{0}} \frac{2 \lambda}{\mathrm{r}} $$

$~$

(iv) Electric field strength due to a circular ring of radius ’ $R$ ’ with uniform distribution of charge ’ $q$ ’ at any general point ’ $\mathrm{P}$ ’ on the axis of the ring.

$$ \mathrm{E}=\frac{1}{4 \pi \varepsilon _{0}} \frac{\mathrm{q} x}{\left(\mathrm{R}^{2}+x^{2}\right)^{3 / 2}} $$

If $x > >\mathrm{R}$

$$ \mathrm{E}=\frac{1}{4 \pi \varepsilon _{0}} \frac{\mathrm{q}}{x^{2}} $$

$~$

Behaves like point charge.

(v) Electric field strength due to a uniformly surface charged, disc of radius ’ $R$ ’ any general point ’ $\mathrm{P}$ ’ on its axis.

$ \mathrm{E}=\frac{\sigma}{2 \varepsilon _{0}}\left[1-\frac{x}{\sqrt{x^{2}+\mathrm{R}^{2}}}\right] $

If $x > >\mathrm{R}$

$ \mathrm{E}=\frac{1}{4 \pi \varepsilon _{0}} \frac{\theta}{x^{2}} $

Disc behaves like point charge.

$~$

If $x > >\mathrm{R}$

$ \mathrm{E}=\frac{\sigma}{2 \varepsilon _{0}} $

Disc behaves like then plane sheet.

Example-(6):

A ring of radius $\mathrm{R}$ with uniformly distributed charge $\mathrm{q}$ is placed in the $y-z$ plane with its centre at the origin. Plot the variation of $\mathrm{E}$ along the axis of the ring and obtain the maximum value of $\mathrm{E}$ and identify the position of the corresponding point.

$~$

Show Answer

Solution:

We have,

$$ \mathrm{E}=\frac{1}{4 \pi \varepsilon _{0}} \frac{\mathrm{q} x}{\left(\mathrm{R}^{2}+x^{2}\right)^{3 / 2}} $$

For maximum field intensity $\frac{\mathrm{dE}}{\mathrm{d} x}=0$.

$$ \begin{aligned} & \Rightarrow x= \pm \frac{\mathrm{R}}{2} \\ & \Rightarrow \mathrm{E} _{\max }=\frac{\mathrm{q}}{6 \sqrt{3} \pi \varepsilon _{0} \mathrm{R}^{3}} \quad\left(\text { Putting } x=\frac{\mathrm{R}}{2}\right) \end{aligned} $$

The plot is an shown.

$~$

Example-7:

A pair of charges, $+\mathrm{q}$ and $-\mathrm{q}$, are placed at $x=-\mathrm{a}$ and $x=+$ a respectively. Plot the veriation of E along the $x$-axis.

$~$

Show Answer

Solution:

The variation is as shown.

$~$

Electric Field Lines

An electric field line is a curve drawn in three dimensions in such a way that the tangent at each ponit on the curve gives the direction of electric field at that point.

The idea of electric lines of force, or the electric field lines, introduced by Micheal Faraday, is a way to visualize electrislatic field geometrically.

The properties of electric field Inies are:

(i) Electric field lines are continuous curves in an electric field; starting from a positive charge and ending on a negative charge.

$~$

(ii) Two electric field lines never intersect.

(iii) Electric field lines do not pass through, they leave, or end, on a charged conductor normally.

(iv) The relative closeness of field lines indicates the relative strength of electric field at different points; crowded lines represent strong field regions while distant field lines correspond to a weak field. Further, if the field lines are uniformly spaced parallel straight lines, the field is uniform.

(v) It is not necessary that a small test charge released at a point in an electric field will move along the field line passing through that ponit. The field lines give the direction of acceleration they do not essentialy give the direction of motion of charge.

$~$ $~$

Electric Dipole

An electric dipole is a pair of equal and opposite charges $[(+q)$ and $(-q)]$, seperated by a (small) distance ’ $2 \mathrm{a}$ ‘. Any molecule in which the centre of the positive charges does not coincide with the centre of negative charges, may be approximated as an electric dipole.

The basic characterstics of a dipole is its electric dipole moment $(\mathbf{p})$. It is defined as the product of the magnitude of either charge and the separation between the charges and is given by

$$ \mathbf{p}=q(2 \mathbf{a}) $$

$~$

Dipole moment is thus a vector quantity; its direction is the direction of the dipole axis, from $(-q)$ to $(+q)$. Its SI Unit in $\mathrm{Cm}$.

Electric field Intensity due to an Electric Dipole

(i) Axial position : The field, on the axis, at a distance ’ $r$ ’ from the centre:

$~$

The electric field on axial position ’ $A$ ’ can be calculated by vector addition of electric field due to $+q$ and $-q$ at point ’ $A$ '

$$ \begin{aligned} & \mathbf{E} _{\text {axial }}=\mathbf{E} _{+\mathrm{q}}+\mathbf{E} _{-\mathrm{q}} \\ & =\frac{1}{4 \pi \varepsilon _{0}} \frac{2 \mathrm{pr}}{\left(\mathrm{r}^{2}-\mathrm{a}^{2}\right)} \end{aligned} $$

For an ideal electric dipole $\mathrm{r} > > 2 \mathrm{a}$

$$ \therefore \mathbf{E} \cong \frac{1}{4 \pi \varepsilon _{0}} \frac{2 \mathbf{p}}{\mathrm{r}^{3}} $$

Thus, Electric field is along the direction of $\mathbf{p}$, for all axial points.

(ii) Equitorial position : The field on the equitorial plane (i.e. plane perperndicular to the axis and passing through its centre) at a distance ’ $r$ ’ from centre is:

$~$

As seen in the diagram above, electric field on equitorial position is only due to the horizontal components of electric fields due to $(+q)$ and $(-q)$.

$$ \begin{aligned} & \therefore \mathbf{E} _{\mathrm{eq}}=2 \mathbf{E} _{+\mathrm{q}} \cos \theta \\ & \mathbf{E} _{\mathrm{eq}}=\frac{1}{4 \pi \varepsilon _{0}} \frac{-\mathbf{p}}{\left(\mathrm{r}^{2}+\mathrm{a}^{2}\right)^{3 / 2}} \end{aligned} $$

For an ideal dipole : $\mathrm{r} > > 2 \mathrm{a}$

$\therefore \mathbf{E} _{\text {eq }} \simeq-\frac{1}{4 \pi \varepsilon _{0}} \frac{\mathbf{p}}{\mathrm{r}^{3}}$

Thus, electric field is in the direction opposite to $\overrightarrow{\mathrm{p}}$ for points on the equitorial plane.

For the same value of the distance ’ $r$ ‘.

$$ \mathrm{E} _{\text {axial }}=2 \mathrm{E} _{\mathrm{eq}} \quad \text { (For a small dipole) } $$

(iii) Field at any general ponit ’ $r$ ‘.

Let ’ $\theta$ ’ be the angle between $\mathbf{p}$ and $\mathbf{r}$. We can now resolve $\mathbf{P}$ into two components; $\mathbf{p} _{11}$ and $\mathbf{p} _{\perp}$, i.e. parallel to $\mathbf{r}$ and perpendicular to $\mathbf{r}$; point ’ $r$ ’ then lies on the axial line of $\mathbf{p} _{11}$ and lies on the equitorial line of $\mathbf{p} _{\perp}$.

$~$

Hence, $\mathrm{E} _{\mathrm{net}}=\sqrt{\mathrm{E} _{11}^{2}+\mathrm{E} _{\perp}^{2}}$ where

$ \mathrm{E} _{11} \simeq \frac{1}{4 \pi \varepsilon _{0}} \frac{2 \mathrm{p} \cos \theta}{\mathrm{r}^{3}} \qquad \qquad \left(\because \mathrm{p} _{11}=\mathrm{p} \cos \theta\right) $

and $\mathrm{E} _{\perp}=\frac{1}{4 \pi \varepsilon _{0}} \frac{\mathrm{p} \cos \theta}{\mathrm{r}^{3}} \qquad \qquad \left(\because \mathrm{p} _{\perp}=\mathrm{p} \sin \theta\right)$

$\therefore \mathrm{E}=\frac{1}{4 \pi \varepsilon _{0}} \frac{\mathrm{p}}{\mathrm{r}} \mathrm{r} \sqrt{1+3 \cos ^{2} \theta}$

This (general) electric field, at any point ’ $r$ ‘, makes an angle $\alpha$ w.r.t. $\mathbf{E} _{11}$; where $\alpha=\tan ^{-1}\left(\frac{1}{2} \tan \theta\right)$

Example-(8):

The electric field, due to a short dipole, at a distance $r$, on the axial line, from its mid point is the same as the electric field at a distance $\mathrm{r}^{\prime}$ on the equitorial line from its mid point. Determine the ratio $\left(\frac{\mathrm{r}}{\mathrm{r}^{\prime}}\right)$.

Show Answer

Solution:

We are given that

$ \mathrm{E} _{\mathrm{axial}(\mathrm{r})}=\mathrm{E} _{\mathrm{eq}\left(\mathrm{r}^{\prime}\right)} $

$\therefore \frac{1}{4 \pi \varepsilon _{0}} \frac{2 \mathrm{p}}{\mathrm{r}^{3}}=\frac{1}{4 \pi \varepsilon _{0}} \frac{\mathrm{p}}{\left(\mathrm{r}^{\prime}\right)^{3}}$

$\Rightarrow \frac{2}{\mathrm{r}^{3}}=\frac{1}{\mathrm{r^{\prime}}^{3}}$

$\therefore\left(\frac{\mathrm{r}}{\mathrm{r}^{\prime}}\right)=2^{1 / 2}$

Electric dipole placed in a uniform external field

$~$

When a dipole is placed in uniform external field, the net translation motion is zero as:

$$ \begin{aligned} & \mathbf{F} _{+q}=-\mathbf{F} _{-q} \\ & \therefore \mathbf{F} _{\mathrm{net}}=0 \end{aligned} $$

However, the forces $\mathbf{F} _{+q}$ and $\mathbf{F} _{-q}$ form a couple. Their turning effect, or torque, is given by

$$ \begin{aligned} & \tau=(\mathrm{qE})(2 \mathrm{a} \sin \theta) \\ & =\mathrm{pE} \sin \theta \end{aligned} $$

where $\mathrm{q} 2 \mathrm{a}=\mathrm{p}$ (electric dipole moment)

The direction of $\boldsymbol{\tau}$ is given by

$$ \boldsymbol{\tau}=\mathbf{p} \times \mathbf{E} $$

Hence, due to this torque, the dipole tries to allign itself parallel to the direction of electric field. This is its position of stable equilibrium.

Special Case : When the dipole is placed in a non-uniform external electric field, the charges of dipole experience unequal forces; therefore, net force on the dipole is not zero. Now dipole has both translation as well as rotational motion.

Example-(9):

In a certain region of space, electric field is along the $+\mathrm{z}$ direction throughout. The magnitude of electric field is, however, not cnostant but increases uniformly along the $+\mathrm{z}$ direction; at the rate of $\left(10^{5} \mathrm{~N} / \mathrm{C}\right)$, per metre. Find the forces and torque, expreirenced by a system having a total dipole moment equal to $10^{-7} \mathrm{~cm}$ along the negative direction.

The dipole moment, for a segment of length $\mathrm{dz}$, is

$ \mathrm{dp}=\mathrm{qdz} $

$~$

The net force, on this elementaldipole, is

$ \mathrm{dF}=\mathrm{q} \frac{\mathrm{dE}}{\mathrm{dz}} \mathrm{dz} $

$\therefore$ The total force is

$ \mathrm{F}=\int \mathrm{q} \frac{\mathrm{dE}}{\mathrm{dz}} \mathrm{dz} $

$ \begin{aligned} & =10^{+5} \int \mathrm{qdz} \\ & =10^{+5}(\mathrm{p})=10^{+5} \times 10^{-7} \mathrm{~N}-10^{-2} \mathrm{~N} \end{aligned} $

This force is directed along the direction of the dipole moment, i.e., the (-ve) direction of $z-$ axis.

$$ \therefore \mathbf{F}=10^{-2}(-\hat{\mathrm{k}}) \mathrm{N} $$

Work done in Rotation of a Dipole

Suppose the dipole is initially kept in a uniform electric field at an angle $\theta_{i} $ ;

Now to turn it to a position at an anode $\theta_{\mathrm{f}} $,

work done by the field is $ \mathrm{W}=\int_{\theta_ {\mathrm{i}}}^{\theta_ {\mathrm{f}}} \tau \mathrm{d} \theta $ .

This work done is stored in the form of electrostatic potential energy. Thus, potential energy of a dipole in uniform electric field will be

$\mathrm{W}=\int_{\theta_{\mathrm{i}}}^{\theta_{\mathrm{f}}} \tau \mathrm{d} \theta$.

For conservative forces, $\mathrm{W}=-\Delta \mathrm{U}$

$\therefore \Delta \mathrm{U}=\mathrm{pE} \cos \theta _{\mathrm{i}}-\mathrm{pE} \cos \theta _{\mathrm{f}}$

We take $\theta _{\mathrm{i}}=\frac{\pi}{2}$, as the reference position. If $\theta _{\mathrm{f}}=\theta$, we then have

$$ \begin{aligned} & \mathrm{U}=-\mathrm{pE} \cos \theta \\ & \mathrm{U}=-\mathbf{p} \cdot \mathbf{E} \end{aligned} $$

$~$

$\therefore$ At $\theta=0^{0}, \mathrm{U}=-\mathrm{pE}=\mathrm{U} _{\text {min }}$

and $\tau=0$

The dipole is in stable equilibrium.

At $\theta=180^{\circ}, \mathrm{U}=\mathrm{U} _{\text {max }}=+\mathrm{pE}$

and $\tau=0$

Here dipole is in unstable equilibrium.

Special Case:

If a dipole is given a small angular displacement ’ $\theta$ ’ about its quilibrium (stable equilibrium) position, the dipole will execute angular S.H.M. with time period.

$ \mathrm{T}=2 \pi \sqrt{\frac{\mathrm{I}}{\mathrm{PE}}} $

Where I is the moment of inertia of the dipole about an axis passing through its centre. This is because the torque will act like a restoring torque:

$ \tau \simeq-\mathrm{pE}(\theta)=\frac{\operatorname{Id}^{2} \theta}{\mathrm{dt}^{2}} $

Electric Flux

The electric flux $\Delta \phi$, of an electric field $\mathbf{E}$, through a small area element $\Delta \mathbf{s}$, is given by

$ \Delta \phi=\mathbf{E} \cdot \Delta \mathbf{s} $

The vector area element $\Delta \mathbf{s}$ is giveny by

$ \Delta \mathbf{s}=\Delta \mathrm{s} \hat{\mathrm{n}} $

$\Delta \mathrm{s}$ is the magnitude of the area element and $\hat{\mathrm{n}}$ is the (outward) normal to the area element.

For an area element of a closed surface, $\hat{n}$ is taken to be the direction of outward normal. For a closed surface, outward flux is taken as positive while inward flux is taken as negative.

$~$

The total flux, ’ $\phi$ ‘, therefore linking with surface ’ $s$ ’ is

$ \phi=\int _{\mathrm{s}} \mathbf{E} \cdot \mathbf{d s} $

The SI unit of electric flux, is $\mathrm{Vm}$ or $\mathrm{NC}^{-1} \mathrm{~m}$.

Gauss Law

The electric flux, through any closed surface ’ $s$ ‘, is $\frac{1}{\varepsilon _{0}}$ times the total charge enclosed by ’s’.

$$ \phi=\oint \mathbf{E} \cdot \mathbf{d s}=\frac{\mathrm{q}}{\varepsilon _{0}} \quad \left[ \text{For a given ’enclosed’ charge } \right] $$

The integral $\oint \mathbf{E}$.ds , known as surface intergral of electric field over a closed surface, is independent of the shape and size of the surface and position of the charge inside it.

• The term q on the right side of Gauss’s law includes the sum of the charges enclosed by the surface. The charges my be located anywhere inside the surface.

• The surface that we choose for application of Gauss’s law is called as the Gaussian surface.

• On the left side of gauss law, for calculation of $\mathbf{E}$, all the charges contribute whether they lie inside or outside the Gaussian surface.

• Gauss law is based on the inverse square dependence on distance contained in the Coulomb’s law. Any violation of Gauss’s law will indicate departure from the inverse square law.

• Gauss law is expecially useful in determining electric field $\mathrm{E}$, when the source distribution has a simple symmetry : like the spherical or planar or cylindrical charge distribution.

(a) Thin infinitely long straight wire of uniform linear charge density ’ $\lambda$ ‘.

$~$

$\mathbf{E}=\frac{\lambda}{2 \pi \varepsilon _{0} \mathrm{r}} \hat{\mathrm{n}}$

where ’ $r$ ’ is the perpendicular distance of the poinnt from the wire and $\hat{n}$ is the radial unit vector in the plane normal to the wire passing through the point.

$~$

(b) Infinite thin non-conducting plane sheet, of uniform surface charge density ’ $\sigma$ ‘. (Non-conducting uniformaly charged plane sheet).

$~$

$\mathbf{E}=\frac{\sigma}{2 \varepsilon _{0}} \hat{n}$

where $\hat{n}$ is a unit vector normal to the plane, outward on either

side. For conducting sheet $\mathbf{E}=\frac{\sigma}{\varepsilon _{0}} \hat{n}$.

$~$

(c) Thin shperical shell of uniform surface charge density ’ $\sigma$ ‘.

$~$

$\mathbf{E}=\frac{1}{4 \pi \varepsilon _{0}} \frac{\mathrm{q}}{\mathrm{r}^{2}} \hat{\mathrm{r}}=\frac{\sigma \mathrm{R}^{2}}{\varepsilon _{0} \mathrm{r}^{2}} \hat{\mathrm{r}}(\mathrm{r} \geq \mathrm{R})$

$=0(\mathrm{r}<\mathrm{R})$

where $r$ is the distance of the point from the centre of the shell and $R$ the radius of the shell.

The electric field inside the shell iszero and outside it behaves as that of a point charge.

(d) Long (Non-conducting) uniformly charged rod/cylinder having uniform volume charged density ’ $\rho$ ‘.

$\mathrm{E}=\frac{\rho \mathrm{R}^{2}}{2 x \varepsilon _{0}}(x \geq \mathrm{R})$

$~$

$=\frac{\rho x}{2 \varepsilon _{0}}(x<\mathrm{R})$

Where $x$ is the distance of point from the axis of rod / cylinder.

$~$

(e) Solid (Non-conducting) sphere of total charge ’ $q$ ’ having volume charge density ’ $\mathrm{f}$ ‘.

$$ \begin{aligned} & \mathrm{E}=\frac{1}{4 \pi \varepsilon _{0}} \frac{\mathrm{q}}{\mathrm{r}^{2}} \hat{\mathrm{r}}=\frac{\rho \mathrm{R}^{3}}{3 \varepsilon _{0} \mathrm{r}^{2}} \hat{\mathrm{r}}(\mathrm{r} \geq \mathrm{R}) \\ & \mathrm{E}=\frac{1}{4 \pi \varepsilon _{0}} \frac{\mathrm{qr}}{\mathrm{R}^{3}} \hat{\mathrm{r}}=\frac{\rho \mathrm{r}}{3 \varepsilon _{0}}(\mathrm{r}<\mathrm{R}) \end{aligned} $$

$~$

For a solid conductor $\mathrm{E}=0(\mathrm{r}<\mathrm{R})$ (conducting solid sphere) inside.

Hence, A conducting solid sphere acts like a spherical shell.

(f) Long uniformaly charged conducting cylinder / rod.

$$ \begin{aligned} & \mathrm{E}=\frac{\sigma \mathrm{R}}{\varepsilon _{0} x}(x \geq \mathrm{R}) \\ & =0(x<\mathrm{R}) \end{aligned} $$

For a conductor, electric field strength at interior points is zero.

$~$

(g) Two infinite thin plates having uniform surface charge density held parallel to each other.

$~$

$\mathbf{E} _{\mathrm{A}}=-\left(\frac{\sigma _{1}+\sigma _{2}}{2 \varepsilon _{0}}\right) \hat{\mathrm{n}}$

$\mathbf{E} _{\mathrm{C}}=-\left(\frac{\sigma _{1}+\sigma _{2}}{2 \varepsilon _{0}}\right) \hat{\mathrm{n}}$

$\mathbf{E} _{\mathrm{B}}=-\left(\frac{\sigma _{1}+\sigma _{2}}{2 \varepsilon _{0}}\right) \hat{\mathrm{n}}$

Special Case : For two plane sheets with equal and opposite charge densities.

$\mathrm{E}=\frac{\sigma}{2 \varepsilon _{0}}$ directed from ‘positive’ plate to negative plate, between the plane sheets.

$\mathrm{E}=0$ outside the plane sheets.

Example-(10):

A point charge $\mathrm{q}$ is placed at a distance $\ell$ from a disc of radius R. Find the electric flux through the surface of disc due to this point charge.

$~$

Show Answer

Solution:

Consider an elemental ring on the disc surface; of radius $x$ and width $\mathrm{d} x$ as shown in the figure.

Area of this strip $=2 \pi x d x$

Electric field ’ $E$ ‘, due to $q$, at this ring is

$$ \mathrm{E}=\frac{1}{4 \pi \varepsilon _{0}} \frac{\mathrm{q}}{\left(x^{2}+\ell^{2}\right)} $$

If $\mathrm{d} \phi$ is the flux passing through this ring, we have,

$$ \begin{aligned} & \mathrm{d} \phi=\text { Eds } \cos \theta \\ & =\frac{1}{4 \pi \varepsilon _{0}} \frac{\mathrm{q}}{\left(x^{2}+\ell^{2}\right)} 2 \pi x \mathrm{~d} x \times \frac{\ell}{\left(x^{2}+\ell^{2}\right)} \\ & =\frac{x \mathrm{q} \ell \mathrm{d} x}{2 \varepsilon _{0}\left(\ell^{2}+x^{2}\right)^{3 / 2}} \end{aligned} $$

Total flux through the disc surface,

$$ \begin{aligned} & \phi=\int _{0}^{\mathrm{R}} \mathrm{d} \phi=\frac{\mathrm{q} \ell}{2 \varepsilon _{0}} \int _{0}^{\mathrm{R}} \frac{x \mathrm{~d} x}{\left(\ell^{2}+x^{2}\right)^{3 / 2}} \\ & \Rightarrow \phi=\frac{\mathrm{q} \ell}{2 \varepsilon _{0}}\left[\frac{1}{\ell}-\frac{1}{\ell^{2}+\mathrm{R}^{2}}\right] \end{aligned} $$

Example-(11):

Consider a ring of radius ’ $r$ ’ on which a charge $q$ has been uniformly distributed. If a shepre of radius $r$ is centred on the circumference of the ring. Find the electric flux passing through the surface of the spere.

$~$

Show Answer

Solution:

It is clear from the geometry of the figure that $\theta=60^{\circ}$. So one third part of the ring lies inside the sphere

hence, enclosed charge is $\frac{q}{3}$ only.

Using Gauss law, electric flux passing through the surface of sphere.

$$ \phi=\frac{1}{3} \frac{\mathrm{q}}{\varepsilon _{0}} $$

Example-(12):

A spherical conducting shell of inner radius $r _{1}$ and outer radius $r _{2}$ has a charge ’ $Q$ ‘. A charge of $q$ is placed at the centre of the shell.

(a) What is the surface charge density on the (i) inner surface (ii) outer surface of the shell?

(b) Write the expression for the electric field at a point $x>\mathrm{r} _{2}$ from the centre of the shell.

$~$

Show Answer

Solution:

(a) (i) Surface charge density on the inner surface of the shell.

$ \sigma _{1}=\frac{-q}{4 \pi r _{1}^{2}} $

(ii) Surface charge density on the outer surface of the shell.

$ \sigma _{2}=\frac{\mathrm{Q}+\mathrm{q}}{4 \pi \mathrm{r} _{2}^{2}} $

(b) Using Gauss’s law

$ \begin{aligned} & \mathrm{E} 4 \pi x^{2}=\frac{\mathrm{Q}+\mathrm{q}}{\varepsilon _{0}} \\ & \mathrm{E}=\frac{1}{4 \pi \varepsilon _{0}} \frac{\mathrm{Q}+\mathrm{q}}{x^{2}} \end{aligned} $

The Conservation Nature of the Electrostatic Field

It is an important property of electrostatic fields that the line integral of electric field,

$ \int _{\mathrm{A}}^{\mathrm{B}} \mathbf{E} \cdot \mathbf{d} \ell $

depends only on the initial and final position and is independent of the path taken in going from Ato B.

$~$

Consider two points A and B in the electrostatic field of a charge ’ $q$ ’ kept at origin.

We then get: $\int _{\mathrm{A}}^{\mathrm{B}} \mathbf{E} \cdot \mathbf{d} \ell=\frac{1}{4 \pi \varepsilon _{0}} \mathrm{q}\left[\frac{1}{\mathrm{r} _{\mathrm{A}}}-\frac{1}{\mathrm{r} _{\mathrm{B}}}\right]$

The line integral is, therefore, seem to be independent of path connecting A and B. We can conclude that line integral of electrostatic field (does not depends upon the curve $\mathrm{L} _{1}$ or $\mathrm{L} _{2}$ ) depends only on the position vectors of the initial and final points.

For a closed curve the initial and final positions are the same. Hence

$ \oint \mathbf{E} \cdot \mathbf{d} \ell=0 $

The line integral of electrostatic field over a closed path is zero; the electrostatic field is conservative in nature.

Consider the work done in displacing a unit positive charge through a distance ’ $\mathrm{d} \ell$ ’ in the field;

$$ \begin{aligned} & \mathrm{d} \omega=\mathbf{F} \cdot \mathrm{d} \ell=\mathrm{q} \mathbf{E} \cdot \mathbf{d} \ell \\ & =1 \times \mathbf{E} \cdot \mathrm{d} \ell=\mathbf{E} \cdot \mathrm{d} \ell \end{aligned} $$

Total work done, in moving a unit positive charge from $\mathrm{A}$ to $\mathrm{B}$, then is

$$ \mathrm{w}=\int _{\mathrm{A}}^{\mathrm{B}} \mathrm{dw}=\int _{\mathrm{A}}^{\mathrm{B}} \mathbf{E} \cdot \mathbf{d} \ell $$

Thus line intergral of electric field represents the amount of work done is moving a unit positive charge, between two given points, in an electrostatic field.

This work done by the elctrostatic force, is path independent, as electrostatic force is conservative in nature. The concept of potential energy exists only in the case of conservative forces. When electrostatic force exists, between two or more point charges within system, we can assign a potential energy $U$ to the system (due to two interaction of the point charges).

Electrostatic Potential Energy

Consider an electrostatic field ’ $\mathrm{E}$ ’ due to a charge $\mathrm{Q}$ placed at the origin. Now, imagine that we brieng a test charge $q$ from a point $R$ to a point $P$ against the repulsive force on it due to charge ’ $Q$ ‘, without any acceleration. (i.e. we apply an external force. $F _{\text {ext }}$ just enough to counter the electrostatic repulsive force F).

Thus, work done by external force in moving a charge q from $\mathrm{R}$ to $\mathrm{P}$.

$$ \mathrm{W} _{\mathrm{RP}}=\int _{\mathrm{R}}^{\mathrm{P}} \mathbf{F} _{\mathrm{ext}} \cdot \mathbf{d r}=\int _{\mathrm{R}}^{\mathrm{P}} \mathrm{Fdr} $$

This work done is against electrostatic repulsive force $\mathrm{F}$ and is path independent. At every point, test charge q possesses certain potential energy, this work done increases its potential energy.

Thus, potential energy difference -

$$ \Delta \mathrm{U}=\mathrm{U} _{\mathrm{P}}-\mathrm{U} _{\mathrm{R}}=\mathrm{w} _{\mathrm{RP}} $$

$~$

Therefore we can define electric potential energy difference between two points as the work required to be done by an external force in moving (without acceleration) charge q from one point to another for electric field of any arbitary charge configuration.

A convenient choice is to have electrostatic potential energy zero at infinity with his choice if we take point ’ $R$ ’ at infinity, we can write.

$$ \mathrm{W} _{\infty \mathrm{P}}=\mathrm{U} _{\mathrm{P}}-\mathrm{U} _{\infty}=\mathrm{U} _{\mathrm{P}}=\frac{1}{4 \pi \varepsilon _{0}} \frac{\mathrm{Qq}}{\mathrm{r}} $$

Thus, potential energy of a charge $q$ at a point (in the presence of field due to any charge configuration) is the work done by the external force (equal and opposite to electrostatic force) in bringing the charge $q$ from infinity to that point.

Electric Potential at a Point

Electric / electrostatic potential ’ $\mathrm{V}$ ’ at any point in a region with electrostatic field is the work done in bringing a unit positive charge (without acceleration) from infinity to that point.

Thus, $\mathrm{V}=\frac{\Delta \mathrm{w}}{\Delta \mathrm{q}}=\frac{\mathrm{U}}{\mathrm{q}}$

$ = \int_{\infty}^{r} E.dl $

$~$

Hence potential $\mathrm{V}$ at ’ $\mathrm{P}$ ’ due to the charge $\mathrm{Q}$ is given by

$ \mathrm{V}(\mathrm{r})=\frac{1}{4 \pi \varepsilon _{0}} \frac{\mathrm{Q}}{\mathrm{r}} $

Important Points

• $\quad$ Electric potential is a scalar quantity, its value may be positive, negative or zero.
• $\quad$ SI Unit of potential is $=\frac{1 \mathrm{Joule}}{1 \mathrm{Coulomb}}$ or $1 \mathrm{~V}=\frac{1 \mathrm{~J}}{1 \mathrm{C}}$.
• Electric potential at a point is also equal to the negative of the work done by the field in taking the unit positive charge from infinity to that point i.e. $\mathrm{V} _{\mathrm{A}}-\mathrm{V} _{\mathrm{B}}=$ work done in moving a unit positive charge from B to A, or $\int _{\mathrm{A}}^{\mathrm{B}} \mathbf{E} \cdot \mathbf{d} \ell=\mathrm{V} _{\mathrm{A}}-\mathrm{V} _{\mathrm{B}}$.

The above result is valid only for electric potential of a point charge. If we wish to find the electric potential due to a continuous charge distribution (charged extended body), we first find the potential due to an element charge ’ $\mathrm{dq}$ ’ on the body and then intergrate the expression over the whole charge distribution.

Electric Potential due to Various Charge Distribution

(i) Electric potential due to an isolated point charge at a point ’ $\mathrm{P}$ ‘.

$~$

$\mathrm{V}=\frac{1}{4 \pi \varepsilon _{0}}$

for $\mathrm{q}>0, \mathrm{~V}+$ ve for $\mathrm{q}<0, \mathrm{~V}-\mathrm{ve}$

(ii) Electric potential due to a system of ’ $n$ ’ point charges at a point ’ $\mathrm{P}$ ‘.

$~$

$\mathrm{V}=\mathrm{V} _{1}+\mathrm{V} _{2}+\mathrm{V} _{3} \ldots+\mathrm{V} _{\mathrm{n}}$

(iii) Electrical potential due to a charged rod of length ’ $\ell$ ’ at a point $\mathrm{P}$ having uniform charge ’ $\theta$ ‘.

$~$

$\mathrm{V}=\frac{1}{4 \pi \varepsilon _{0}} \frac{\theta}{\ell} \ell n \frac{(\mathrm{r}+\ell)}{\mathrm{r}}$

(iv) Electric potential due to a charged ring having uniform charge ’ $\mathrm{Q}$ ’ at its centre having radius $\mathrm{R}$.

$~$

$\mathrm{V}=\frac{1}{4 \pi \varepsilon _{0}} \frac{\mathrm{Q}}{\mathrm{R}}$

(v) Electric potential due to a uniformally charged sing on the axis of the ring at a point ’ $\mathrm{P}$ ‘.

$~$

$\mathrm{V}=\frac{1}{4 \pi \varepsilon _{0}} \frac{\mathrm{Q}}{\sqrt{\mathrm{R}^{2}+x^{2}}}$

(vi) Electric field due to a circular disc of radius $\mathrm{R}$ with uniform charge density ’ $\sigma$ ’ at a point ’ $\mathrm{P}$ ‘.

$~$

$\mathrm{V}=\frac{\sigma}{2 \varepsilon _{0}}\left(\sqrt{x^{2}+\mathrm{R}^{2}-x}\right)$

(vii) Electric field due to a solid spere of radius ’ $R$ ’ with volume charge density ’ $\rho$ ’ and $V=\frac{\rho R^{3}}{3 \varepsilon _{0}} \frac{1}{r}$ for, $\mathrm{r}>\mathrm{R}$.

$$ \mathrm{V}=\frac{\rho \mathrm{R}^{2}}{6 \varepsilon _{0}}\left(\frac{3-\mathrm{r}^{2}}{\mathrm{R}^{2}}\right) \text { for } \mathrm{r} \leq \mathrm{R} $$

$~$

Example-(13):

A metal sphere $0.30 \mathrm{~m}$ in radius is positively charged with $2 \mu \mathrm{C}$. Find the potential at (a) $1 \mathrm{~m}$ from the centre of sphere (b) a position on the sphere (c) the centre of sphere.

Show Answer

Solution:

(a) $\mathrm{V}=\frac{1}{4 \pi \varepsilon _{0}} \frac{\mathrm{q}}{\mathrm{r}}$

$ =\frac{9 \times 10^{9} \times 2 \times 10^{-6}}{(1)} $

$=18 \mathrm{KV}$

(b) $\mathrm{V}=\frac{1}{4 \pi \varepsilon _{0}} \frac{\mathrm{q}}{\mathrm{r}}$

$=\frac{9 \times 10^{9} \times 2 \times 10^{-6}}{0.3}$

$=60 \mathrm{KV}$

(c) Potential at the centre of sphere is same as on the surface.

Example-(14):

Eight changed water droplets, each with a radius $1 \mathrm{~mm}$ and a charge of $10^{-10} \mathrm{C}$, coalesce to form a single drop. Calculate the potential of the bigger drop.

Show Answer

Solution:

Let $\mathrm{R}$ and $\mathrm{r}$ be the radii of the bigger and small droplet.

$\Rightarrow$ Volume of 8 droplets $=$ Volume of bigger drop

$$ \begin{aligned} & =8 \frac{4}{3} \pi r^{3}=\frac{4}{3} \pi r^{3} \\ & \Rightarrow \mathrm{R}=2 \mathrm{r} \\ & =2 \times 10^{-3} \mathrm{~m} \end{aligned} $$

As charge is conserved total charge $\mathrm{Q}=8 \times 10^{-10} \mathrm{C}$

Potential of bigger drop $\mathrm{V}=\frac{1}{4 \pi \varepsilon _{0}} \frac{\mathrm{Q}}{\mathrm{R}}$

$\mathrm{V}=\frac{9 \times 10^{9} \times 8 \times 10^{-10}}{2 \times 10^{-3}}$

$=3.6 \times 10^{4} \mathrm{~V}$

Relation between Electric Field Strength and Electric Potential

Consider two surfaces $A$ and $B$ having potential $V$ and $V+d V$, where $d V$ is the change in potential in the direction of $\mathbf{E}$.

Let a unit positive charge is moved from $\mathrm{A}$ to $\mathrm{P}$ in the direction of $\mathbf{E}$; then work done in this process is $|\mathbf{E}| \mathrm{d}$.

$~$

Thus, $\mid \mathbf{E} \mathrm{d} \ell=\mathrm{V}-(\mathrm{V}+\mathrm{dV})$

or $\quad|\mathbf{E}|=-\frac{\mathrm{dV}}{\mathrm{d} \ell}=-\frac{\mathrm{dV}}{\mathrm{dr}}$ as $(\mathrm{d} \ell=\mathrm{dr})$

or $\quad \int d V=-\int E d r$

$\Rightarrow \mathrm{V} _{\mathrm{B}}-\mathrm{V} _{\mathrm{A}}=-\int _{\mathrm{r} _{\mathrm{A}}}^{\mathrm{r} _{\mathrm{B}}} \mathrm{Edr}$

Important Conclusion

• Electric field always pionts in the direction in which the potenital decreases steepest.
• Magnitude of the component of electric field (in any direction) is given by the change in the manitude of potential per unit displacement. In otehr words magnitude of component of electric field is numerically equal to the potential gradient.

$$ \text { i.e. } \frac{N}{C}=\frac{V}{m} $$

Equipotential Surface

Equipotential surface is a surface over which potential has a constant value.

Hence, no work is required to move a test charge on the equipotential surface. Thus, electric field, at every point, is normal to the equipotential surafce passing through that point.

Qualitatively, we can determine the magnitude and direction of electric field by understanding the following points about equipotential surfaces:

(i) The density of equipotential surfaces gives an idea about the magnitude of electric field. Higher the density, larger the field strength.

(ii) The direction of electric field is normal to equipotential at every point and points in the direction of decreasing potential.

$~$

Example-(15):

The potential of an electrostatic field is given by $\mathrm{V}=2 x^{2}$. Determine the field strength at the point $(2 \mathrm{~m}, 0$, $3 \mathrm{~m})$.

Show Answer

Solution:

We known that $\mathrm{E} _{x}=-\frac{\delta \mathrm{v}}{\delta x} ; \mathrm{E} _{\mathrm{y}}=-\frac{\delta \mathrm{v}}{\delta \mathrm{y}} ; \mathrm{E} _{\mathrm{z}}=-\frac{\delta \mathrm{v}}{\delta \mathrm{z}}$

$\therefore \mathrm{E} x=-\frac{\mathrm{d}}{\mathrm{d} x}\left(2 x^{2}\right)=-4 x$

Also $\mathrm{E} _{\mathrm{y}}=\mathrm{E} _{\mathrm{z}}=0$

$\therefore$ In vector form $\mathbf{E}=-4 x \hat{\mathrm{i}}$

Hence $\mathbf{E}$ at $(2 \mathrm{~m}, 0,3 \mathrm{~m})$ is given by

$$ E=-4 \times 2 \hat{i}=-(8 \hat{i}) \frac{N}{C} $$

Example-(16):

The figure shows field lines and equipotentials (dashed lines)

(i) Find the external work done to move a $-2 \mu \mathrm{C}$ charge, at constant speed from $\mathrm{A}$ to $\mathrm{B}$, along the path shown.

(ii) Also, determine the work done by the electric field.

$~$

Show Answer

Solution:

Since electric field is conservative, therefore, work done by electric force is independent of path.

Using work-energy theorem, we have

$$ \mathrm{W} _{\mathrm{ext}}+\mathrm{W} _{\text {electric }}=\Delta \mathrm{k} $$

The there is no change in speed, $\Delta \mathrm{k}=0$

$$ \begin{aligned} & \mathrm{W} _{\text {ext }}=-\mathrm{W} _{\text {electric }} \\ & \mathrm{W} _{\text {ext }}=\mathrm{q}\left(\mathrm{V} _{\mathrm{B}}-\mathrm{V} _{\mathrm{A}}\right)=-2 \times 10^{-6}(-25-(-15)) \\ &=20 \mu \mathrm{J} \\ & \Rightarrow \mathrm{W} _{\text {electric }}=-20 \mu \mathrm{J} \end{aligned} $$

Electric Potential due to an Electric Dipole

Consider an electric dipole consisting of two charges $q$ and $-q$ separated by small distance ’ $2 a$ ‘. Electric potential at any point ’ $\mathrm{P}$ ’ at a distance ’ $\mathrm{r}$ ’ from the centre is given by:

$~$

$ \mathrm{V} = \frac{1}{4\pi \epsilon _0} \frac{p \cos \theta }{4 \pi r^2}$

Where ’ $\theta$ ’ is the angle between $\mathbf{p}$ and $\mathbf{r}$.

as $p \cos \theta=\mathbf{p} . \mathbf{r}$

$$ \mathrm{V}=\frac{1}{4 \pi \varepsilon _{0}} \frac{\mathbf{p} \cdot \mathbf{r}}{\mathrm{r}^{2}}(\mathrm{r} > > \mathrm{a}) $$

if $\theta=0^{0}$ or $\pi$, the potential an dipole axis is given by:

$$ \mathrm{V}= \pm \frac{1}{4 \pi \varepsilon _{0}} \frac{\mathrm{p}}{\mathrm{r}^{2}} \quad\quad\left(+\operatorname{sign} \text { for } \theta=0^{0},-\operatorname{sign} \text { for }=180^{\circ}\right) $$

If $\theta=\frac{\pi}{2}$, the potential, in the equitorial plane, is ‘zero’.

Important Conclusion

The potential due to a dipole, depends not just on distance r, but also an the angle ’ $\theta$ ’ between $\mathbf{p}$ and $\mathbf{r}$. It, however, has axially summetry about point ’ $p$ ‘.

Electric Potential Energy of a System of ’ $n$ ’ Point Charges

The electric potential energy of a system of ’ $n$ ’ point charges is the work done to assemble this sytem of charges by bringing them in from an infinite distance.

$$ \begin{gathered} \mathrm{U}=\frac{1}{2} \frac{1}{4 \pi \varepsilon _{0}} \sum _{\mathrm{j}=1}^{\mathrm{n}} \sum _{\mathrm{i}=1}^{\mathrm{n}} \frac{\mathrm{q} _{\mathrm{i}} \mathrm{q} _{\mathrm{j}}}{\mathrm{r} _{\mathrm{ij}}} \\ \\ \mathrm{i} \neq \mathrm{j} \end{gathered} $$

If $\mathrm{U}$ is $+\mathrm{ve}$, it indicates work is done by the external agency to assemble these charges. This work done is stored in form of potential energy. If $U$ is $-v e$, it indicates work is done by the elctrostatic forces to assemble these charges. This means potential energy is released.

Potential Energy in an External Field

• Potential energy of a single point charge in external field at ’ $\mathrm{r}$ ‘, $\mathrm{U}=$ work done on $\mathrm{q}$ against the field.

$$ \mathrm{U}=\mathrm{q} \mathrm{V}(\mathrm{r}) $$

$ \qquad$ Where $\mathrm{V}(\mathrm{r})$ is the external potential at the ponit ’ $\mathrm{r}$ ‘.

• Potential energy of a system of two charges in an external field,

$ \qquad \mathrm{U}=$ workdone on $\mathrm{q} _{1}$ against the field + work done on $\mathrm{q} _{2}$ against the field + workdone in bringing $\mathrm{q} _{2}$ in field on $\mathrm{q} _{1}. $ (Mutual interaction energy)

$ \qquad \mathrm{U}=\mathrm{q} _{1} \mathrm{~V}\left(\mathrm{r} _{1}\right)+\mathrm{q} _{2} \mathrm{~V}\left(\mathrm{r} _{2}\right)+\frac{1}{4 \pi \varepsilon _{0}} \frac{\mathrm{q} _{1} \mathrm{q} _{2}}{\mathrm{r} _{12}}$

• Potential energy of an electric dipole havnig dipole moment ’ $p$ ’ and charges separated by a (small) distance ’ $2 \mathrm{a}$ '

$ \qquad \mathrm{U}=\mathrm{q}(\mathrm{V}(+\mathrm{q})-\mathrm{V}(-\mathrm{q}))-\frac{1}{4 \pi \varepsilon _{0}} \frac{\mathrm{q}^{2}}{2 \mathrm{a}}$

$ \qquad \mathrm{V}=-\mathbf{p} \cdot \mathbf{E}-\frac{1}{4 \pi \varepsilon _{0}} \frac{\mathrm{q}^{2}}{2 \mathrm{a}}$ where $-\mathbf{p} \cdot \mathbf{E}=[\mathrm{q}[\mathrm{V}(+\mathrm{q})-\mathrm{V}(-\mathrm{q})]=-\mathrm{qE} 2 \mathrm{a} \cos \theta=-\mathrm{pE} \cos \theta]$

Example-(17):

Two charges $7 \mu \mathrm{C}$ and $-2 \mu \mathrm{C}$ are placed at $(-9 \mathrm{~cm}, 0,0)$ and $(9 \mathrm{~cm}, 0,0)$ repsectively. Suppose that system of charges is now placed in an external field, $\mathrm{E}=\mathrm{A} \frac{1}{\mathrm{r}^{2}}, \mathrm{~A}=9 \times 10^{5} \mathrm{~cm}^{2}$ what would be the electrostatic energy of configuration?

Show Answer

Solution:

Mutual interaction energy of two charges will be

$$ \begin{aligned} \mathrm{U} _{1}= & \frac{1}{4 \pi \varepsilon _{0}} \frac{\mathrm{q} _{1} \mathrm{q} _{2}}{\mathrm{r}} \\ = & \frac{9 \times 10^{9} \times 7 \times(-2) \times 10^{-12}}{0.18} \\ = & -0.7 \mathrm{~J} \end{aligned} $$

The additional energy of interaction in external field will be

$$ \begin{aligned} \mathrm{U} _{2} = & \mathrm{q} _{1} \mathrm{~V}\left(\mathrm{r} _{1}\right)+\mathrm{q} _{2} \mathrm{~V}\left(\mathrm{r} _{2}\right) \\ = & \mathrm{A} \frac{7 \mu \mathrm{C}}{0.09 \mathrm{~m}}+\mathrm{A} \frac{-2 \mu \mathrm{C}}{0.09 \mathrm{~m}}=50 \mathrm{~J} \\ \mathrm{U} _{\text {net }}= & \mathrm{U} _{1}+\mathrm{U} _{2} \\ = & (50-0.7)=49.3 \mathrm{~J} \end{aligned} $$

Electrostatics of Conductors

In conductors, electric charges (elections) are free move throughout the volume of such bodies. When there is not net motion of free charges within the conductor, the conductor is in electrostatic equilibrium. The following are the properties of a conductor in electrostatic equilibrium.

(i) Electrostatic field is zero everywhere inside the conductor.

(ii) Net volume charge density inside a conductor is zero. Any excess charge must reside at the surface.

(iii) At the surface of a charged conductor an electrostatic field must be normal to the surface at every point.

(iv) Electrostatic potential is constant throughout the volume of conductor and has the same value (inside) as an its surafce.

(v) Electric field at the surface of a charged conductor is:

$$ \mathbf{E}=\frac{\sigma}{\varepsilon _{0}} \hat{n} $$

Where $\sigma$ is the surface charge density and $\hat{n}$ is a unit vector normal to the surface of conductor along the outward direction.

Electrostatic Shielding

If there is a cavity inside the conductor or it is hollow, the field strength inside the cavity equals zero, whatever is the field outside the conductor.

Hence whatever be the charge and field configuration outside, any cavity in a conductor remains sheilded from outside electric infulence. This is known as electrostatic sheilding.

Important Points

• If a charge ’ $q$ ’ is kept in the cavity, then a charge ( $-q$ ) will be induced on the inner surface and $(+q)$ will be induced on the outer surface of the conductor.
• If a charge ’ $q$ ’ is kept inside the cavity of a conductor, and conductor is given a charge $Q$, then a charge ( $-q$ ) will be induced on the inner surface and total charge on the outer surface will be $(q+Q)$

$~$

Example-(18):

A small sphere, of radius $r _{1}$, having a charge $q _{1}$ is, enclosed by a shpherical shell of radius $r _{2}$ and charge $q _{2}$. Show that if $q _{1}$ is positive, charge will necessarily flow from sphere to the shell (when the two are connected by a wire) no matter what the charge $\mathrm{q} _{2}$ on the shell is.

$~$

Show Answer

Solution:

Electric potential at the point ’ $A$ '

$$ \mathrm{V} _{\mathrm{A}}=\frac{1}{4 \pi \varepsilon _{0}} \frac{\mathrm{q} _{1}}{\mathrm{r} _{1}}+\frac{1}{4 \pi \varepsilon _{0}} \frac{\mathrm{q} _{2}}{\mathrm{r} _{2}} $$

Electric potential at the piont ’ $\mathrm{B}$ '

$$ \begin{array}{r} \mathrm{V} _{\mathrm{B}}=\frac{1}{4 \pi \varepsilon _{0}} \frac{\mathrm{q} _{1}}{\mathrm{r} _{2}}+\frac{1}{4 \pi \varepsilon _{0}} \frac{\mathrm{q} _{2}}{\mathrm{r} _{2}^{2}} \\ \Rightarrow \mathrm{V} _{\mathrm{A}}-\mathrm{V} _{\mathrm{B}}=\frac{1}{4 \pi \varepsilon _{0}} \mathrm{q} _{1}\left(\frac{1}{\mathrm{r} _{1}}-\frac{1}{\mathrm{r} _{2}}\right) \end{array} $$

The p.d $\left(V _{A}-V _{B}\right)$ is independent of $q _{2}$. If $q _{1}$ is positive, then $\left(V _{A}-V _{B}\right)$ is positive.

Hence inner small sphere is always at higher potential and if connected by a wire, charge will immediately flow from the sphere to the shell.

Dielectrics

Dielectrics are (generally) non-conducting substances. Glass, plastic, mica, oil etc. are examples of dielectrics. The molecules of a dielectric, substance may be ‘polar’ or ’non-polar’ in nature.

When a dielectric material is placed in an external field, the elctric porperties get modified. In case of nonpolar molecules (zero permanent dipole moment of molecules) the positive and negative charges of the molecules experience electrostatic forces in opposite direction. Therefore, the centre of two charges get separated from each other. The molecules thus acquire an induced electric dipole moment in the direction of the field.

Let us now consider a dielectric whose molecules are polar (i.e. having a permanent electric dipole moments). In the absence of any external electric field, the individual dipoles (molecules) are oriented at random; hence the net dipole moment of the dielectric is zero. When such a dielectric is placed in electric field, the molecules tend to allign themselves parallel to the field due to the torque experienced by them in the field. Thus, a net dipole moment, is produced in the field.

We can thus conclude that, an electric field, induced in a dielectric, (polar or non-polar) a net dipole moment along the direction of the field. This phenomenon is known as dielectric polarisation or polarisation of matter.

Behaviour of dielectric slab is an electric field

Suppose a slab of dielectric material is placed in a uniform electric field $\mathrm{E} _{0}$ created says, by two parallel plates. Its molecules (electric dipoles) will try to orient themselves in the direction of the field. The net effect is the appearance of a negative charge on the face of the slab infront of the positive plate and equal positive charge on the face infront of the negative plate. The net charge in the interior is zero. These induced charges produce their own electric field $\mathrm{E}$ in the direction opposite to external field $\mathbf{E} _{\mathbf{0}}$. Hence we conclude that when a dielectric is placed in an external field, the field in medium is reduced.

$~$

The induced dipole moment, per unit volume, is denoted by the vector $\mathbf{P}$. For a linear dielectric

$ \mathbf{P}=x _{\mathrm{e}} \mathbf{E} $

Where $x _{\mathrm{e}}$ is a constant called the electric suspectibility.

Capacitors

A capacitor is a system of two conductors separated by an insulator usually, in practice, the two conductors have charges $+\mathrm{Q}$ and $-\mathrm{Q}$, and a potential difference ’ $\mathrm{V}$ ’ between them. [Even a single conductor can be used as capacitor by assuming the other conductor at infinity].

The conductors may be charged by connecting them to the two terminals of a battery.

Relation between $\mathbf{Q}$ and $\mathrm{V}$

The electric field in the region between the conductors is proportional to the charge $Q$. That is, if the charge on the capacitor is double the electric field will also be doubled at every point.

$~$

Consequently, $\mathrm{V}$ is also propotional to $\mathrm{Q}$.

Mathematically $\mathrm{Q} \propto \mathrm{V}$

i.e. $\mathrm{Q}=\mathrm{CV}, \mathrm{C}$ is a constant of proportionality known as the Capacitance of the capacitor.

The capacitance $\mathrm{C}$ depends only on the geometrical configuration (size, shape, and separation) of the system of two conductors.

The SI unit of capacitance is 1 farad $=\frac{1 \text { coulomb }}{1 \text { volt }}$ or $1 \mathrm{~F}=\frac{1 \mathrm{C}}{1 \mathrm{~V}}$

The equation shows that for large $\mathrm{C}, \mathrm{V}$ is small for a given ’ $\mathrm{Q}$ ‘. This means a capacitor, with large capacitance can hold large amount of charge, at a relatively small p.d. between the conductors.

The proportionality, between $\mathrm{Q}$ and $\mathrm{V}$, is true only untill $\mathrm{V}$ becomes so large that dielectric breakdown occurs. If dielectric or insulating, medium breakdown occurs the capacitor no longer behaves like a perfect one. So it is important to keep ’ $V$ ’ below the breakdown potential of the capacitor.

Capacitors are mainly used to store charge and electrical energy. A capacitor with fixed capacitance is symbollically shown as

, while the one with variable capacitance is shown as .

Depending upon their geometry, capacitors are categorised as:

(i) Parallel plate capacitors

(ii) Cylindrical capacitors

(iii) Spherical capacitors

Capacitance of a Parallel Plate Capacitor

$~$

A parallel plate capacitor consists of two conducting long plates parallel to each other and separated by a small distance which is small compared to linear dimensions of plates.

The electric field between the plates is given by $\mathrm{E}=\frac{\sigma}{\varepsilon _{0}}$

Where $\sigma$ is the surface charge density on either plate.

If $A$ is the area of each plate and $Q$ charge on each plate then.

$$ \mathrm{E}=\frac{\mathrm{Q}}{\varepsilon _{0} \mathrm{~A}} $$

As potential difference between the plates in given by $\mathrm{V}=\mathrm{Ed}=\frac{\mathrm{Q}}{\varepsilon _{0} \mathrm{~A}} \mathrm{~d}$

Capacitance of parallel plate is given by $\mathrm{C}=\frac{\mathrm{Q}}{\mathrm{V}}=\frac{\varepsilon _{0} \mathrm{~A}}{\mathrm{~d}}$ in vaccum(or air).

If the space between the plates is filled by a medium, of dielectric constant $k$, the capacitance.

$$ \mathrm{C} _{\mathrm{m}}=\mathrm{kC} _{0} $$

Energy Stored in a Capacitor

The energy of a charged capacitor is measured by the total work done in charging the capacitor to a given potential.

$$ \begin{aligned} & \mathrm{dw}=\mathrm{vdq} \\ & \mathrm{w}=\int \mathrm{dw}=\int _{0}^{\mathrm{Q}} \frac{\mathrm{q}}{\mathrm{c}} \mathrm{dq} \end{aligned} $$

This work done is stored as electrostatic potential energy in the field of capactor and is given by.

$$ \begin{align*} & \mathrm{U}=\frac{1}{2} \frac{\mathrm{Q}^{2}}{\mathrm{C}} \\ & =\frac{1}{2} \mathrm{CV}^{2} \\ & =\frac{1}{2} \mathrm{QV} \qquad \qquad \qquad\qquad \ldots . \ldots {1} \end{align*} $$

$~$

Energy density is the energy stored per unit volume. For the parallel place capacitor, we have $u=U _{d} A d$. From this, we get

$$ \begin{equation*} \mathrm{U} _{\mathrm{d}}=\frac{1}{2} \varepsilon _{0} \mathrm{E}^{2} \qquad \qquad \qquad\qquad \ldots . \ldots {2} \end{equation*} $$

Combination of Capacitors

When there is a combination of capacitors in a circuit, we can replace the combination with an equivalent capacitor.

An equivalent capacitor is a single capacitor that has the same value of capacitance as that of actual combination.

(i) $\quad$ Series Combination: All capacitors will have same charge but different potential difference across them. As $\mathrm{Q}$ in constant, $\mathrm{V}=\mathrm{V} _{1}+\mathrm{V} _{2}+\mathrm{V} _{3} \ldots \ldots \ldots+\mathrm{V} _{\mathrm{n}}$. Thus, for capacitors connected $\mathrm{Q}$ in series; we get

$$ \frac{1}{\mathrm{C} _{\mathrm{eq}}}=\frac{1}{\mathrm{C} _{1}}+\frac{1}{\mathrm{C} _{2}}+\ldots \ldots \frac{1}{\mathrm{C} _{\mathrm{n}}} $$

• In series combination the equivalent capacitance is always less than the least value of the capacitances, given in the electrical network.

$~$

• As charge in constant, $\mathrm{V} \propto \frac{1}{\mathrm{C}}$, hence in series combination the smallest capacitor has the maximum potential difference between its plates.

• Energy stored in series combination is given by

$$ \mathrm{U} _{\text {combination }}=\frac{1}{2} \frac{\mathrm{Q}^{2}}{\mathrm{C} _{\mathrm{eq}}} $$

(ii) $\quad$ Parallel Combination: All the capacitors will have same potential difference but different amount of charges. As $V$ is constant, $Q=Q _{1}+Q _{2}+Q _{3} \ldots . . Q _{n}$.

Thus for ’ $n$ ’ capacitors connected in parallel, $\mathrm{C} _{\mathrm{eq}}=\mathrm{C} _{1}+\mathrm{C} _{2}+\ldots . . \mathrm{C} _{\mathrm{n}}$.

• In parallel combination equivalent capacitance is always greater than the largest value of capacitance given in combination.
• As $\mathrm{V}$ is constant, $\mathrm{Q} \propto \mathrm{C}$ hence maximum change appears across the capacitors of largest value.

$~$

• Energy stored in parallel combination is given by:

$$ \mathrm{U} _{\text {combination }}=\frac{1}{2} \mathrm{C} _{\mathrm{eq}} \mathrm{V}^{2} $$

Redistribution of Charges

Consider two insulated capacitors A and B of capacitance $C _{1}$ and $C _{2}$, carrying charges $\mathrm{q} _{1}$ and $\mathrm{q} _{2}$ respectively. Let $V _{1}$ and $V _{2}$ be their respetive potentials. When the two capacitors are joined by a thin metal wire, the positive changes will flow from higher potential to lower potential till both the capacitors accquire some

common potential. Thus charges are redistributed but total charge $q=q _{1}+q _{2}$ remains constant.

$$ \begin{aligned} & \mathrm{V} _{\text {common }}=\frac{\mathrm{Q} _{\text {net }}}{\mathrm{C} _{\mathrm{net}}}=\frac{\mathrm{q} _{1}+\mathrm{q} _{2}}{\mathrm{C} _{1}+\mathrm{C} _{2}} \\ & =\frac{\mathrm{C} _{1} \mathrm{~V} _{1}+\mathrm{C} _{2} \mathrm{~V} _{2}}{\mathrm{C} _{1}+\mathrm{C} _{2}} \end{aligned} $$

Before connection:

$$ \mathrm{U} _{\mathrm{i}}=\frac{1}{2} \mathrm{C} _{1} \mathrm{~V} _{1}^{2}+\frac{1}{2} \mathrm{C} _{2} \mathrm{~V} _{1}^{2} $$

After connection:

$$ \mathrm{U} _{\mathrm{f}}=\frac{1}{2}\left(\mathrm{C} _{\mathrm{net}}\right) \mathrm{V} _{\mathrm{common}}^{2} $$

Hence there is a loss in electrostatic energy in the form of heat.

$$ \left|\mathrm{U} _{\mathrm{f}}-\mathrm{U} _{\mathrm{i}}\right|=\Delta \mathrm{U}=\frac{1}{2} \frac{\mathrm{C} _{1} \mathrm{C} _{2}}{\mathrm{C} _{1}+\mathrm{C} _{2}}\left(\mathrm{~V} _{1}-\mathrm{V} _{2}\right)^{2} $$

$~$

Before connection

Change on $\mathrm{C} _{1}$ after connection $\mathrm{q} _{1}^{1}=\mathrm{C} _{1} \mathrm{~V} _{\text {common }}=\left(\frac{\mathrm{C} _{1}}{\mathrm{C} _{1}+\mathrm{C} _{2}}\right) \mathrm{q}$

Change on $\mathrm{C} _{2}$ after connection $\mathrm{q} _{2}^{1}=\mathrm{C} _{2} \mathrm{~V} _{\text {common }}=\left(\frac{\mathrm{C} _{2}}{\mathrm{C} _{1}+\mathrm{C} _{2}}\right) \mathrm{q}$

‘Dielectric’ in a Parallel Plate Capacitor

Consider a parallel plate capacitor, having two large plates, each of area A, separated by a distance ’ $\mathrm{d}$ ’ in vacuum.

Situation-1: A dielectric, of constant $k$, is inserted between the plates of a fully charged capacitor, after the supply was disconnected.

$~$

Let the thickness of dielectric slab be $\mathrm{t}(\mathrm{t}<\mathrm{d})$.

As capacitor is fully changed and battery disconnected, charge on each plate,

$$ \begin{aligned} & \mathrm{q}=\mathrm{q} _{0}=\text { Constant } \\ & \mathrm{V}=\mathrm{E} _{0}(\mathrm{~d}-\mathrm{t})+\mathrm{E} _{\mathrm{mt}} \\ & \Rightarrow \mathrm{V}=\mathrm{V} _{0}\left(1-\frac{\mathrm{t}}{\mathrm{d}}+\frac{\mathrm{t}}{\mathrm{kd}}\right) \end{aligned} $$

The potential between the plates decreases from its value $\mathrm{V} _{0}$ (in vacuum or air) as $\mathrm{C}=\frac{\mathrm{q}}{\mathrm{V}}=\frac{\mathrm{q} _{0}}{\mathrm{~V}}$ it comes

out, $\mathrm{C}=\frac{\mathrm{C} _{0}}{\left(1-\frac{\mathrm{t}}{\mathrm{d}}+\frac{\mathrm{t}}{\mathrm{kd}}\right)}$, the capacitance increases from its value $\mathrm{C} _{0}$ (in vacuum or air)

$$ \mathrm{U}=\frac{1}{2} \frac{\mathrm{q} _{0}^{2}}{\mathrm{C}}=\mathrm{U} _{0}\left(1-\frac{\mathrm{t}}{\mathrm{d}}+\frac{\mathrm{t}}{\mathrm{kd}}\right) $$

The energy stored in the field of capacitor decreases from its original value $\mathrm{U} _{0}$ (in vacuum or air) If $\mathrm{t}=\mathrm{d}$, (the dielectric completely fills the space between the plates).

We have, $\mathrm{q}=\mathrm{q} _{0}$

and $\mathrm{V}=\frac{\mathrm{V} _{0}}{\mathrm{k}}$

$\therefore \mathrm{C}=\mathrm{C} _{0} \mathrm{k}$

and $\mathrm{U}=\frac{\mathrm{U} _{0}}{\mathrm{k}}$

Situation-2: A dielectric slab of constant $k$ is inserted between the plates of the capacitor while the voltage supply remain connected (during charging of the capacitor).

$\mathrm{V}=\mathrm{V} _{0}=$ constant

$\mathrm{C}=\mathrm{kC} _{0}$, capacitance increases from its original value $\mathrm{C} _{0}$ (in vacuum or air).

$\mathrm{q}-=\mathrm{kq} _{0}$, change storage increases from its original value $\mathrm{q} _{0}$ (in vacuum or air).

$$ \mathrm{U}=\frac{1}{2} \mathrm{CV}^{2}=\mathrm{kV} _{0} $$

Energy stored from its original value $\mathrm{V} _{0}$ (in air or vacuum).

Example-(19):

Find the equivalent capacitance between points $\mathrm{A}$ and $\mathrm{B}$.

$~$

Show Answer

Solution:

Let the equivalent capacitance is $x$, then net capacitance after CD point will be ’ $2 x$ ’ because every capaciatnce becomes 2 times as compared A and B.

$~$

$$ \begin{aligned} & \Rightarrow x=\frac{(\mathrm{C}+2 x) \mathrm{C}}{\mathrm{C}+2 x+\mathrm{C}} \\ & \Rightarrow x=\frac{\mathrm{C}}{\sqrt{2}} \end{aligned} $$

Example-(20):

In a parallel plate capacitor of capacitance $C$, the plate $A$ carries a positive charge and the plate $B$ carries a negative charge. The potential difference between the plates is $\mathrm{V} _{0}$. If the plate $\mathrm{A}$ is given an additional charge ’ $+Q$ ‘, then determine the potential difference between the plates.

Show Answer

Solution:

It is important to note that if the capacitance of the parallel plate capacitor is ’ $C$ ‘, $q+Q$ then the capacitance of each isolated plate will be $2 \mathrm{C}$.

Thus, the potential of plate A will be,

$$ \mathrm{V} _{\mathrm{A}}=\frac{\mathrm{q}+\mathrm{Q}}{2 \mathrm{C}} $$

and that of $B$ will be, $V _{B}=\frac{-q}{2 C}$.

$~$

The p.d. between the plates, now will be:

$$ \begin{aligned} & \Rightarrow \mathrm{V}=\mathrm{V} _{\mathrm{A}}-\mathrm{V} _{\mathrm{B}} \\ & \mathrm{V}=\frac{\mathrm{Q}+\mathrm{q}}{2 \mathrm{C}}-\frac{\mathrm{q}}{2 \mathrm{C}} \\ & \mathrm{V}=\frac{\mathrm{q}}{\mathrm{C}}+\frac{\mathrm{Q}}{2 \mathrm{C}} \end{aligned} $$

Therefore $\mathrm{V}=\mathrm{V} _{0}+\frac{\mathrm{Q}}{2 \mathrm{C}}$ (Since $\mathrm{q}=\mathrm{CV} _{0}$ given)

Example-(21):

Find out capacitance between $\mathrm{A}$ and $\mathrm{B}$ if three dielectric slabs of dielectric constant $\mathrm{K} _{1}$, of area $\mathrm{A} _{1}$ and thickness $\mathrm{d} _{1}, \mathrm{~K} _{2}$ of area $\mathrm{A} _{2}$ and thickness $\mathrm{d} _{2}$ and $\mathrm{K} _{3}$ of area $\mathrm{A} _{3}$ and thickness $\mathrm{d} _{3}$ are inserted between the plates of parallel plate capacitor of plate area A as shown in figure. (Given distance between the two plates $\left.\mathrm{d} _{1}=\mathrm{d} _{2}+\mathrm{d} _{3}\right)$

$~$

Show Answer

Solution:

The capacitor behaves like three individual capacitors $\mathrm{C} _{1}, \mathrm{C} _{2}$ and $\mathrm{C} _{3}$ such that $\mathrm{C} _{2} \& \mathrm{C} _{3}$ are in series and their resultant is in parallel to $\mathrm{C} _{1}$.

Thus, $\mathrm{C} _{\mathrm{eq}}=\mathrm{C} _{1}+\frac{\mathrm{C} _{2} \mathrm{C} _{3}}{\mathrm{C} _{2}+\mathrm{C} _{3}}$

$ =\frac{\mathrm{A} _{1} \mathrm{~K} _{1} \varepsilon _{0}}{\mathrm{~d} _{2}+\mathrm{d} _{3}}+\frac{\frac{\mathrm{A} _{2} \mathrm{~K} _{2} \varepsilon _{0}}{\mathrm{~d} _{2}} \frac{\mathrm{A} _{3} \mathrm{~K} _{3} \varepsilon _{0}}{\mathrm{~d} _{3}}}{\frac{\mathrm{A} _{2} \mathrm{~K} _{2} \varepsilon _{0}}{\mathrm{~d} _{2}} \frac{\mathrm{A} _{3} \mathrm{~K} _{3} \varepsilon _{0}}{\mathrm{~d} _{3}}} $

$\Rightarrow \mathrm{C} _{\mathrm{eq}}=\frac{\mathrm{A} _{1} \mathrm{~K} _{1} \varepsilon _{0}}{\mathrm{~d} _{2}+\mathrm{d} _{3}}+\frac{\mathrm{A} _{2} \mathrm{~K} _{2} \mathrm{~K} _{2} \varepsilon _{0}}{\mathrm{~K} _{2} \mathrm{~d} _{3}+\mathrm{k} _{3} \mathrm{~d} _{2}}$

ADDITIONAL EXAMPLES

1. The region between two concentric spheres of radii ’ $a$ ’ and ’ $b$ ‘, respectively (see figure), has volume charge density $\rho=\frac{A}{r}$, where $A$ is a constant and $r$ is the distance from the centre. At the centre of the sphere is a point charge $Q$. The value of A such that the electric field in the region between the sphere will be constant is:

$~$

(1) $\frac{\mathrm{Q}}{2 \pi\left(\mathrm{b}^{2}-\mathrm{a}^{2}\right)}$

(2) $\frac{2 Q}{\pi\left(b^{2}-a^{2}\right)}$

(3) $\frac{2 \mathrm{Q}}{\pi \mathrm{a}^{2}}$

(4) $\frac{\mathrm{Q}}{2 \pi \mathrm{a}^{2}}$

Correct option (4)

Show Answer

Solution:

Let $\mathrm{E}$ be the electric field at any point distance $\mathrm{r}$ from the centre. We than have $\mathrm{E}=$ electric field, at a distance $\mathrm{r}$, due to a point charge $\mathrm{Q}$ ’ where, $Q^{\prime}=Q+$ charge in the region between $r=a$ and $r=r$

$=\mathrm{Q}+\int _{0}^{\mathrm{r}} \rho \cdot\left(4 \pi \mathrm{r}^{2} \mathrm{dr}\right)$

$=\mathrm{Q}+2 \pi\left(\mathrm{r}^{2}-\mathrm{a}^{2}\right) \mathrm{A}$

$\therefore \mathrm{E}=\frac{1}{4 \pi \varepsilon _{0}}\left[\frac{\mathrm{Q}+2 \pi\left(\mathrm{r}^{2}-\mathrm{a}^{2}\right) \mathrm{A}}{\mathrm{r}^{2}}\right]$

This has to be independent of r. We, therefore, have

$ \begin{aligned} & \frac{\mathrm{Q}}{\mathrm{r}^{2}}=\left(\frac{2 \pi \mathrm{a}^{2} \mathrm{~A}}{\mathrm{r}^{2}}\right)=0 \\ & \therefore \mathrm{A}=\frac{\mathrm{Q}}{2 \pi \mathrm{a}^{2}} \end{aligned} $

2. A long cylinderal shell carries positive surface charge $\sigma$ in the upper half and negative surface charge $-\sigma$ in the lower half. The electric field lines around the cylinder will look like figure given in:

(Figures are schematic and not drawn to scale)

$~$

$~$

Correct option (3)

(IIT 2015)

Show Answer

Solution:

The correct figure is figure 3 . This is because electric field lines start from positive charges and terminate on negative charges.

3. Assume that an electric field $\mathrm{E}=30 x$ i exists in space. Then the potential difference $\mathrm{V} _{\mathrm{A}}-\mathrm{V} _{0}$, where $\mathrm{V} _{0}$ is the potential at the origin and $\mathrm{V} _{\mathrm{A}}$ the potential at $x=2 \mathrm{~m}$ is: (1) $-80 \mathrm{~V}$ (2) $80 \mathrm{~V}$ (3) $150 \mathrm{~V}$ (4) $-150 \mathrm{~V}$

Correct option (1)

(IIT 2014)

Show Answer

Solution:

We have, $\mathrm{E} _{x}=-\frac{\mathrm{dv}}{\mathrm{d} x}$

$\therefore \mathrm{dv}=-\mathrm{E} _{x} \mathrm{~d} x$

$\therefore \mathrm{V} _{\mathrm{A}}-\mathrm{V} _{0}=-\int _{0}^{2} 30 x^{2} \mathrm{~d} x$

$$ =-30\left|\frac{x^{3}}{3}\right| _{0}^{2}=-80 \mathrm{~V} $$

4. A uniformly charged solid sphere of radius $\mathrm{R}$ has potential $\mathrm{V} _{0}$ (measured with respect to $\infty$ ) on its surface. For this sphere the equipotential surfaces with potentials $\frac{3 \mathrm{~V} _{0}}{2}, \frac{5 \mathrm{~V} _{0}}{4}, \frac{3 \mathrm{~V} _{0}}{4}$ and $\frac{\mathrm{V} _{0}}{4}$ have radius $\mathrm{R} _{1}, \mathrm{R} _{2}, \mathrm{R} _{3}$ and $\mathrm{R} _{4}$ respectively. Then

(1) $\mathrm{R} _{1}=0$ and $\mathrm{R} _{2}<\left(\mathrm{R} _{4}-\mathrm{R} _{3}\right)$

(2) $2 \mathrm{R}<\mathrm{R} _{4}$

(3) $\mathrm{R} _{1}=0$ and $\mathrm{R} _{2}>\left(\mathrm{R} _{4}-\mathrm{R} _{3}\right)$

Correct option (1)

Show Answer

Solution:

Let the sphere have a charge $+Q$. Then Potential

(i) On the surface $\mathrm{V} _{0}=\frac{\mathrm{KQ}}{\mathrm{R}}$

(ii) inside $(0<r<R) V _{1}=\frac{K Q}{2 R^{3}}\left(3 R^{2}-r^{2}\right)$

$\therefore \frac{3 \mathrm{~V} _{0}}{2}=\frac{3}{2} \frac{\mathrm{KQ}}{\mathrm{R}}$

(At the cenre $r=0$ )

Therefore, radius of equi-potential at the centre of sphere is zero. $R _{1}=0$

Let $R _{2}$ be the radius of second equi-potential surface

$\therefore \frac{5 \mathrm{~V} _{0}}{4}=\frac{\mathrm{KQ}}{2 \mathrm{R} _{3}}\left(3 \mathrm{R}^{2}-\mathrm{R} _{2}^{2}\right)$

$\therefore 2 \mathrm{R} _{2}^{2}=\mathrm{R}^{2} \Rightarrow \mathrm{R} _{2}=\frac{2}{\sqrt{2}}$

Let $R _{3}$ be the radius of third equi-potential surface

$\therefore \frac{3 \mathrm{~V} _{0}}{4}=\frac{\mathrm{KQ}}{\mathrm{R} _{3}}$

$$ \frac{3 \mathrm{KQ}}{4 \mathrm{R}}=\frac{\mathrm{KQ}}{\mathrm{R} _{3}} \Rightarrow \mathrm{R} _{3}=\frac{4 \mathrm{R}}{3} $$

Let $R _{4}$ be the radius of third equi-potential surface

$\frac{\mathrm{V} _{0}}{4}=\frac{\mathrm{KQ}}{\mathrm{R} _{4}}$

$\therefore \frac{\mathrm{KQ}}{4 \mathrm{R}}=\frac{\mathrm{KQ}}{\mathrm{R} _{4}} \Rightarrow \mathrm{R} _{4}=4 \mathrm{R}$

$\therefore \quad \mathrm{R} _{4}-\mathrm{R} _{3}=4 \mathrm{R}-\frac{4 \mathrm{R}}{3}=\frac{8 \mathrm{R}}{3}=2.66 \mathrm{R}$

and $\quad \mathrm{R} _{2}=\frac{\mathrm{R}}{1.414}<\mathrm{R}$

$\therefore \mathrm{R} _{2}<\left(\mathrm{R} _{4}-\mathrm{R} _{3}\right)$

5. A combination of capacitor is set up as shown in the figure. The magnitude of the electric field, due to a point charge $Q$ (having a charge equal to the sum of the charges on the $4 \mu \mathrm{F}$ and $9 \mu \mathrm{F}$ capacitors), at a point distant $30 \mathrm{~m}$ from it, would equal:

(1) $360 \mathrm{~N} / \mathrm{C}$

(2) $420 \mathrm{~N} / \mathrm{C}$

(3) $480 \mathrm{~N} / \mathrm{C}$

(4) $240 \mathrm{~N} / \mathrm{C}$

$~$

Correct option (2)

(IIT 2015)

Show Answer

Solution:

The equivalent capacitance of the top row is $3 \mu \mathrm{F}$.

$\therefore \mathrm{Q}=3 \times 8 \mu \mathrm{C}=24 \mu \mathrm{C}$

This charge, on the $4 \mu \mathrm{F}$ capacitor, gets divided, in the ratio 1:3 between the $3 \mu \mathrm{F}$ and $9 \mu \mathrm{F}$ capacitor.

Hence $\mathrm{Q} _{4}=\frac{3}{4} \times 24 \mu \mathrm{C}$

$ =18 \mu \mathrm{C} $

$~$

$\therefore$ The charges on $4 \mu \mathrm{F}$ and $9 \mu \mathrm{F}$ capacitros are charges $24 \mu \mathrm{C}$ and $18 \mu \mathrm{C}$ respectively.

Hence, $\mathrm{E}=\frac{9 \times 10^{9} \times(24+18) \mu \mathrm{C}}{30 \times 30}=420 \mathrm{~N} / \mathrm{C}$

6. In the given circuit, charge $\mathrm{Q} _{2}$ on the $2 \mu \mathrm{F}$ capacitor changes as $\mathrm{C}$ is varied from $1 \mu \mathrm{F}$ to $3 \mu \mathrm{F} . \mathrm{Q} _{2}$ as a function of ’ $\mathrm{C}$ ’ is given properly by : (figures are drawn schematically and are not to scale).

$~$

$~$

Correct option (1)

(IIT 2015)

Show Answer

Solution:

For $\mathrm{C}=1 \mu \mathrm{F}$, equivalent capacitance $=\frac{3}{4} \mu \mathrm{F}$

$\therefore \mathrm{Q} _{2}=\frac{3}{4} \mathrm{~V} \times \frac{2}{3} \mu \mathrm{C}=\frac{\mathrm{V}}{2} \mu \mathrm{C}$

For $\mathrm{C}=2 \mu \mathrm{F}$, Equivalent capacitance $=\frac{5}{6} \mu \mathrm{F}$

$\therefore \mathrm{Q} _{2}=\frac{5}{6} \mathrm{~V} \times \frac{2}{3} \mu \mathrm{C}=\frac{5 \mathrm{~V}}{8} \mu \mathrm{C}$

For $\mathrm{C}=3 \mu \mathrm{F}$, Equivalent capacitance $=\frac{3}{2} \mu \mathrm{F}$

$\therefore \mathrm{Q} _{2}=\frac{3}{2} \mathrm{~V} \times \frac{2}{3} \mu \mathrm{C}=\mathrm{V} \mu \mathrm{C}$

$\therefore \mathrm{Q} _{2}$ iscreases (non-linearly) with increasing $\mathrm{C}$.

Also change in $\mathrm{Q} _{2}$, in changing $\mathrm{C}$ from $2 \mu \mathrm{F}$ to $3 \mu \mathrm{F}$ is more than that in changing $\mathrm{C}$ from $1 \mu \mathrm{F}$ to $3 \mu \mathrm{F}$.

$\therefore$ Graph 1 is the correct graph.

7. A parallel plate capacitor is made of two circular plates separated by a distance of $5 \mathrm{~mm}$ and with a dielectric of dielectric constant 2.2 between them. When the electric field in the dielectric is $3 \times 10^{4} \mathrm{~V} /$ $\mathrm{m}$, the charge density of the positive plate will be close to:

(1) $3 \times 10^{4} \mathrm{C} / \mathrm{m}^{2}$

(2) $6 \times 10^{4} \mathrm{C} / \mathrm{m}^{2}$

(3) $6 \times 10^{-7} \mathrm{C} / \mathrm{m}^{2}$

(4) $3 \times 10^{-7} \mathrm{C} / \mathrm{m}^{2}$

Correct option (3)

(IIT 2014)

Show Answer

Solution:

Electric field (between the plates of the parallel plate capacitor) $\mathrm{E}=\frac{\sigma}{\mathrm{K} \varepsilon _{0}}$

$\therefore \sigma=\mathrm{EK} \varepsilon _{0}=3 \times 10^{4} \times 2.2 \times 8.85 \times 10^{-12} \mathrm{C} / \mathrm{m}^{2}$

$\cong 6 \times 10^{-7} \mathrm{C} / \mathrm{m}^{2}$

8. A charge $\mathrm{Q}$ is uniformly distributed over a long rod $\mathrm{AB}$ of length $\mathrm{L}$ as shown in the figure. The electric potential at the print $\mathrm{O}$ lying at a distance $\mathrm{L}$ from the end $\mathrm{A}$ is:

$~$

(1) $\frac{3 \mathrm{Q}}{4 \pi \varepsilon _{0} \mathrm{~L}}$

(2) $\frac{Q}{4 \pi \varepsilon_{0} \mathrm{~L} ~ \mathrm{In} ~ 2}$

(3) $\frac{\mathrm{Q} \ell n 2}{4 \pi \varepsilon _{0} \mathrm{~L}}$

(4) $\frac{\mathrm{Q}}{8 \pi \varepsilon _{0} \mathrm{~L}}$

$\frac{\mathrm{Q}}{4 \pi \varepsilon _{0} \mathrm{~L} \ell n 2}$

Correct option (3)

(IIT 2013)

Show Answer

Solution:

$~$

Potential at $\mathrm{O}$, due to the small element $\mathrm{d} x$, is $\mathrm{dV}=\frac{1}{4 \pi \varepsilon _{0}}\left[\left(\frac{\mathrm{Q}}{\mathrm{L}}\right) \mathrm{d} x\right] \cdot \frac{1}{x}$

$$ \begin{aligned} & \therefore \mathrm{V}=\frac{\mathrm{Q}}{4 \pi \varepsilon _{0} \mathrm{~L}} \int _{x=\mathrm{L}}^{x=2 \mathrm{~L}} \frac{\mathrm{d} x}{x} \\ & =\frac{\mathrm{Q}(l n ~ 2)}{4 \pi \varepsilon _{0} \mathrm{~L}} \end{aligned} $$

9. Two capacitors $\mathrm{C} _{1}$ and $\mathrm{C} _{2}$ are charged to $120 \mathrm{~V}$ and $200 \mathrm{~V}$ respectively. It is found that by connecting them together the potential on each one can be made zero.Then:

(1) $3 \mathrm{C} _{1}+5 \mathrm{C} _{2}$

(2) $3 \mathrm{C} _{1}+5 \mathrm{C} _{2}=0$

(3) $9 \mathrm{C} _{1}+4 \mathrm{C} _{2}$

(4) $5 \mathrm{C} _{1}+3 \mathrm{C} _{2}$

Correct option (1)

(IIT 2013)

Show Answer

Solution:

We need to have the charges on the two, to be equal in magnitude.

$\therefore 120 \mathrm{C} _{1}=200 \mathrm{C} _{2}$

$\therefore 3 \mathrm{C} _{1}=5 \mathrm{C} _{2}$

$~$

10. Two charges, each equal to $\mathrm{q}$, are kept at $x=-\mathrm{a}$ and $x=\mathrm{a}$ on the $x$-axis. A particle of mass $\mathrm{m}$ and charge $\mathrm{q} _{0}=\frac{\mathrm{q}}{2}$ is placed at the origin. If charge $\mathrm{q} _{0}$ is given a small displacement $(\mathrm{y} < < \mathrm{a})$ along the $\mathrm{y}$-axis, the net force acting on the particle is proportional to: (1) $-\mathrm{y}$ (2) $\frac{1}{y}$ (3) $-\frac{1}{y}$ (4) $\mathrm{y}$

Correct option (4)

(IIT 2013)

Show Answer

Solution:

We see from the figure, that $\mathrm{F} _{\text {net }}=2 \mathrm{~F} \cos \theta$

$$ =2 F \cdot \frac{y}{\sqrt{a^{2}+y^{2}}} $$

$~$

$$ \begin{aligned} & =2 \cdot \frac{1}{4 \pi \varepsilon _{0}} \frac{\left(\mathrm{q} \cdot \frac{\mathrm{q}}{2}\right)}{\left(\mathrm{a}^{2}+\mathrm{y}^{2}\right)} \frac{\mathrm{y}}{\sqrt{\mathrm{a}^{2}+\mathrm{y}^{2}}} \\ & =\frac{\mathrm{q}^{2}}{4 \pi \varepsilon _{0}} \cdot \frac{1}{\left(\mathrm{a}^{2}+\mathrm{y}^{2}\right)^{3 / 2}} \cdot \mathrm{y} \end{aligned} $$

For $\mathrm{y} < < \mathrm{a}$, we, therefore, have

$$ \begin{aligned} & \mathrm{F} _{\mathrm{net}} \simeq\left(\frac{\mathrm{q}^{2}}{4 \pi \varepsilon _{0} \mathrm{a}^{3}}\right) \mathrm{y} \\ & \therefore \mathrm{F} _{\text {net }} \alpha \mathrm{y} \end{aligned} $$

PROBLEMS FOR PRACTICE

1. Two equal point charges $\mathrm{Q}=+\sqrt{2} \mu \mathrm{C}$ are placed at each of the two opposite corners of a square and equal point charge $q$ at each of the other two corners. What must be the value of $q$ so that the resultant force on $\mathrm{Q}$ is zero?

Show Answer

Answer: $0.5 \mu \mathrm{C}$

2. Two identical charged spheres are suspended in air by strings of equal lengths and make an angle of $30^{0}$ with each other. When suspended in a liquid of density $0.8 \mathrm{~g} / \mathrm{cm}^{3}$ the angle remains the same. What is the dielectric constant of the liquid? (Density of the material of the spheres $=1.6 \mathrm{~g} / \mathrm{cm}^{3}$ ).

Show Answer

Answer: $\mathrm{K}=2$

3. $\mathrm{ABC}$ is an equilateral triangle of side $10 \mathrm{~m}$ and $\mathrm{D}$ is the midpoint of $\mathrm{BC}$. Charges of $+100,-100$ and $+75 \mu \mathrm{C}$ are placed at $\mathrm{B}, \mathrm{C}$, and $\mathrm{D}$ respectively. Find the force on $\mathrm{a}+1 \mu \mathrm{C}$ charge placed at $\mathrm{A}$.

Show Answer

Answer: $9 \sqrt{2} \times 10^{-3} \mathrm{~N}$, at $45^{\circ}$ to a line parallel to $\mathrm{BC}$

4. There is a uniform electric field $\mathrm{E}$ between two parallel plates. An electron moving perpendicular to the field enters with velocity $\mathrm{V}$. Assuming the force of gravity to be negligible, derive an expression for the vertical displacement of the electron when emerging from the plates. Charge of the electron in e and mass $\mathrm{m}$.

$~$

Show Answer

Answer: $\left[\mathrm{y}=\frac{\mathrm{eE}}{2 \mathrm{mv}^{2}} x^{2}, \delta=\frac{\mathrm{eE}}{2 \mathrm{mv}^{2}} \ell^{2}\right]$

5. Two point charges of $+16 \mu \mathrm{C}$ and $-9 \mu \mathrm{C}$ are placed $8 \mathrm{~cm}$ apart in air. Determine the position of the ponit at which the resultant electric field is zero.

Show Answer

Answer: $24 \mathrm{~cm}$ to the right of $-9 \mu \mathrm{C}$ charge

6. Three point charges $\mathrm{q}, 2 \mathrm{q}$ and $8 \mathrm{q}$ are placed on a $9 \mathrm{~cm}$ long straight line. Find the positions where the charges should be placed such that potential energy of this sytem is minimum. In this situation, what will be the electric field at the position of the change $q$ due to the other two charges.

[Hint: For minimum potential energy the charges of maximum magnitude should be farthest apart. Such that $\frac{\mathrm{dU}}{\mathrm{dr}}=0$.

$~$

$\mathrm{U}$ will be minimum for $\mathrm{r}=6 \mathrm{~cm}$. The electric field at $\mathrm{q}$ within situation is $\mathrm{E}=0$.]

7. Three point charges of $1 \mathrm{C}, 2 \mathrm{C}$ and $3 \mathrm{C}$ are placed at the corners of an equilateral triangle of side $1 \mathrm{~m}$. Calculate the work required to move these charges to the corners of a smaller equilateral triangle of side $0.5 \mathrm{~m}$.

Show Answer

Answer: $99 \times 10^{9} \mathrm{~J}$

[Hint: Work required is stored as the additional potential energy of the system.]

8. A change $\mathrm{Q}$ is distributed over two concentric hollow spheres of radii $\mathrm{r}$ and $\mathrm{R}(>\mathrm{r})$ such that the surface charge densities are equal. Find the potential at the common centre.

Show Answer

Answer: $V=\frac{\mathrm{Q}}{4 \pi \varepsilon _{0}}\left(\frac{\mathrm{R}+\mathrm{r}}{\mathrm{r}^{2}+\mathrm{R}^{2}}\right)$

9. Two electric dipoles of moments $p _{1}$ and $p _{2}$ are in straight line show that the potential energy of each in the presence of other is $-\frac{1}{4 \pi \varepsilon _{0}} \frac{\mathrm{p} _{1} \mathrm{p} _{2}}{\mathrm{r} _{3}}$, $\mathrm{r}$ is the distance between the dipoles ( $\mathrm{r}$ is much greater than the length of the dipole).

10. Two charges $\mathrm{q} _{1}=+2 \times 10^{-8} \mathrm{C}$ and $\mathrm{q} _{2}=-0.4 \times 10^{-8} \mathrm{C}$ are placed $60 \mathrm{~cm}$ apart as shown. Athird charge $\mathrm{q} _{3}=$ $+0.2 \times 10^{-8} \mathrm{C}$ is moved along the arc of a circle of radius $80 \mathrm{~cm}$ from $\mathrm{C}$ to $\mathrm{D}$. Find the percentage change in the potential energy of $\mathrm{q} _{3}$.

$~$

Show Answer

Answer: $76 \%$

11. In the figure given, the electric field $\mathrm{E}=100 \mathrm{~N} / \mathrm{C}$ is shown. The side of the cube is $20 \mathrm{~cm}$. Find the nature of the charge which is inside the cube and the magnitude of the charge given $\varepsilon _{0}=8.85 \times 10^{-12} \mathrm{C}^{2} / \mathrm{Nm}^{2}$.

$~$

Show Answer

Answer: $\mathrm{q}=+7.08 \times 10^{-11} \mathrm{C}$

12. In a region of space, the electric field is given by $\mathbf{E}=8 \hat{i}+4 \hat{j}+3 \hat{k}$. Calculate the electric flux through a surface of area 100 units in $x-y$ plane.

Show Answer

Answer: 300 Units

13. Electric field in the given figure is directed along $+x$ direction and given by $\mathrm{E} _{x}=5 \mathrm{~A} _{x}+2 \mathrm{~B}$. Where $\mathrm{E}$ is in $\mathrm{N} / \mathrm{C}$ and $x$ is in metre, $\mathrm{A}$ and $\mathrm{B}$ are constants with dimensions. Taking $\mathrm{A}=10 \mathrm{~N} / \mathrm{Cm}$ and B 5 N/C. Calculate

(i) The electric flux through the cube.

(ii) Net charge enlosed within the cube.

$~$

Given $\varepsilon _{0}=8.85 \times 10^{-12} \mathrm{C}^{2} / \mathrm{Nm}^{2}$

Show Answer

Answer: $\phi _{\text {net }}=0.05 \mathrm{~N} / \mathrm{Cm}^{2}, \mathrm{q}=44.25 \times 10^{-14} \mathrm{C}$

14. In the given circuit, if the ponit $p$ is connected to earth and a potential of $+1200 \mathrm{~V}$ is given to point a, calculate the charges accquired by each of the capacitors and the potential of the point $\mathrm{C}$ ?

$~$

Show Answer

Answer: Charge on $\mathrm{C} _{1}=2.4 \times 10^{-3} \mathrm{C}$

Charge on $\mathrm{C} _{2}=1.6 \times 10^{-3} \mathrm{C}$

Charge on $\mathrm{C} _{3}=0.8 \times 10^{-3} \mathrm{C}$

$ \text { Potential at } \mathrm{C}=400 \mathrm{~V} $

15. In the given network of capacitor, find the effective capacitance betwen the points P and Q. Given, $\mathrm{C} _{1}=\mathrm{C} _{2}=\mathrm{C} _{3}=4 \mu \mathrm{F}$ and $\mathrm{C} _{5}=5 \mu \mathrm{F}$. If a $10 \mathrm{~V}$ battery be connected across $\mathrm{P}$ and $\mathrm{Q}$, what will be the charges on the capacitors?

$~$

Hint: This is a wheatsone bridge arrangement. The bridge formed by capacitors is balanced $\mathrm{R} \& \mathrm{~S}$ are at same potential. Charges on $\mathrm{C} _{1}, \mathrm{C} _{2}, \mathrm{C} _{3}$ and $\mathrm{C} _{4}$ are $20 \mu \mathrm{C}$ and on $\mathrm{C} _{5}=0, \mathrm{C} _{\text {eq }}=4 \mu \mathrm{F}$.

16. A20 $\mu \mathrm{F}$ capacitor charged to $100 \mathrm{~V}$ is connected in parallel to a $10 \mu \mathrm{F}$ capacitor charged to 100 V. Find the loss in energy.

Show Answer

Answer: $2.70 \mathrm{~J}$

17. Two capacitors $25 \mu \mathrm{F}$ and $100 \mu \mathrm{F}$, connected in sereis are charged to $120 \mathrm{~V}$ battery. The battery is then removed. The capactitors are now connected in parallel. Find the (i) p.d. across each and (ii) energy loss in the process.

Show Answer

Answer: p.d. across each $=38.4 \mathrm{~V}$, Energy loss $=0.0518 \mathrm{~J}$

18. A parallel plate capacitor contains one mica sheet of thickness $\mathrm{d} _{1}=1 \times 10^{-3} \mathrm{~m}$ and one fibre sheet of thickness $\mathrm{d} _{2}=0.5 \times 10^{-3} \mathrm{~m}$. Th dielectric constant of mica and fibre are 8 and 2.5 respectively. Fibre breaksdown in an electric field of $6.4 \times 10^{6} \mathrm{~V} / \mathrm{m}$. What maximum voltage can be applied to the capacitor.

Show Answer

Answer: $5200 \mathrm{~V}$

19. A $2 \mu \mathrm{F}$ parallel plate capactor with dielectric slab $(\mathrm{K}=5)$ between the plates is charged to $100 \mathrm{~V}$ and then isolated. What will be the p.d. if the dielectric be removed? How much work will be done in removing the dielectric?

Show Answer

Answer: (i) $500 \mathrm{~V}$

(ii) $0.20 \mathrm{~J}$

20. Find the equivalent capacitance between the given combination.

$~$

Show Answer

Answer: $30 \mu \mathrm{F}$

$~$

Question Bank

Key Learning Points

1. There are two types of charges in nature, like charges repel and unlike charges attract.

2. Electric charges has three basic properties; quantisation, additivity and conservation.

3. Coulomb’s Law: The mutual electrostatic force between two point charges (stationary) is directly propotional to the product $\mathrm{q} _{1} \mathrm{q} _{2}$ and inversely propotional to the square of the distance separeting them. $\mathrm{F}=\frac{\mathrm{Kq} _{1} \mathrm{q} _{2}}{\mathrm{r}^{2}}$ where $\mathrm{K}$ is constant of proportionality, having value in $\mathrm{SI} 9 \times 10^{9} \mathrm{Nm}^{2} \mathrm{C}^{-2}$ in air or vacuum. Also $\mathrm{K}=\frac{1}{4 \pi \varepsilon _{0}}, \varepsilon _{0}$ is called absolute permittivity of the free space, air or vaccum.

$ \varepsilon _{0}=8.85 \times 10^{-12} \mathrm{C}^{2} \mathrm{~N}^{-1} \mathrm{~m}^{-2} $

In vector form, coulamb’s law is written as

$ \mathbf{F} _{12}=\frac{1}{4 \pi \varepsilon _{0}} \frac{\mathrm{q} _{1} \mathrm{q} _{2}}{\mathrm{r} _{12}^{2}} \hat{\mathrm{r}} _{21}=-\mathbf{F} _{21} $

4. Electrostatic forces in medium decreases by a factor of dielectric constant of medium (k)

$$ \frac{\mathrm{F} _{\text {vacuum }}}{\mathrm{F} _{\text {medium }}}=\mathrm{k} $$

5. We can use superposition principle for computing the net force on any charge due to any configuration of charges.

$$ \mathbf{F} _{1}=\frac{1}{4 \pi \varepsilon _{0}} \mathrm{q} _{1} \sum _{\mathrm{i}=2}^{\mathrm{n}} \frac{\mathrm{q} _{\mathrm{i}} \cdot \mathrm{q} _{\mathrm{j}}}{\mathrm{r} _{\mathrm{ij}}{ }^{2}} \hat{\mathrm{r}} _{\mathrm{ji}} $$

6. Electric field strength $\overrightarrow{\mathrm{E}}$ due to a source charge ’ $\mathrm{Q}$ ’ is independent of test charge $\mathrm{q} _{0}$ and is given by

$$ \mathbf{E}=\frac{1}{4 \pi \varepsilon _{0}} \frac{\mathrm{Q}}{\mathrm{r}^{2}} \hat{\mathrm{r}} $$

It is radially outwards from $\mathrm{Q}$, if $\mathrm{Q}$ is positive and radially inwords if $\mathrm{Q}$ is negative. Like Coulamb’s law, electric field also satisfies superposition principle.

7. From the knowledge of electric field strength $\overrightarrow{\mathrm{E}}$ at any point $\overrightarrow{\mathrm{r}}$, we can easily calculate the magnitude and direction of force experienced by any charge $\mathrm{q} _{0}$ held at that point i.e.

$$ \mathbf{F}=\mathrm{q} _{0} \mathbf{E} $$

8. An electrostatic field line is a curve drawn in such a way that tanget at every point on the curve gives the direction of field at that point.

9. Electric field intensity at a point is equal to the number of field lines crossing normally a unit area around that point or the magnitude of field is represented by the density of field lines around that point.

10. Electric field strength of an electric dipole

(i) On axial line of electric dipole at a distance ’ $r$ ’ from centre given is by:

$$ \begin{aligned} & \mathbf{E}=\frac{1}{4 \pi \varepsilon _{0}} \frac{2 \mathrm{pr}}{\left(\mathrm{r}^{2}-\mathrm{a}^{2}\right)^{2}} \\ & =\frac{1}{4 \pi \varepsilon _{0}} \frac{2 \mathbf{p}}{\mathrm{r}^{3}} \text { for } \mathrm{r} > > \mathrm{a} \end{aligned} $$

(ii) On equitorial plane at a distance ’ $r$ ’ fromthe centre is:

$$ \begin{aligned} & \mathbf{E}=\frac{-\mathbf{p}}{4 \pi \varepsilon _{0}} \frac{1}{\left(\mathrm{a}^{2}+\mathrm{r}^{2}\right)^{3 / 2}} \\ & =\frac{-\mathbf{p}}{4 \pi \varepsilon _{0}} \frac{1}{\mathrm{r}^{3}} \text { for } \mathrm{r} > > \mathrm{a} \end{aligned} $$

Thus, the electric field due to a dipole falls off at large distances, at a much faster rate $\left(\alpha \frac{1}{r^{3}}\right)$ then the electric field due to a single charge $\left(\alpha \frac{1}{\mathrm{r}^{2}}\right)$.

11. Electric field due to a single point charge has a spherically symmetry whereas that due to an electric dipole has cylindrical symmetry.

12. In a uniform electric field $\mathrm{E}$, a dipole experiences a torque $\tau$ given by $\boldsymbol{\tau}=\mathbf{p} \times \mathbf{E}$, but experiences no net force.

13. Amount of work done in rotating the dipole through a small angle is given by $\mathrm{w}=-\mathrm{pE} \cos \theta _{\mathrm{f}}+\mathrm{pE} \cos \theta _{\mathrm{i}}$. Which is stored in the form of electrostatic potential energy in the field of electric dipole.

14. The flux $\Delta \phi$ of electric field $\mathrm{E}$ through a small are element $\Delta \mathrm{s}$ is given by

$$ \Delta \phi=\mathrm{E} \cdot \Delta \mathbf{s}=\mathrm{E} \Delta \mathrm{s} \cos \theta $$

Where $\theta$ is the angle between $\mathbf{E}$ and $\hat{\mathrm{n}}$ (to the area element). $\hat{\mathrm{n}}$ is taken as outward normal by convention to calculate the total electric flux through any surface, we integrate.

$$ \phi=\int _{\mathrm{s}} \mathbf{E} . \mathbf{d s} $$

15. Gauss Law: The flux of electric field through any closed surface $\mathrm{s}$ is $\frac{1}{\varepsilon _{0}}$ times the total charge enclosed within the gaussian surface.

$$ \phi=\frac{\mathrm{q} _{\text {net }}}{\varepsilon _{0}} $$

It is used for symmetric charge configuration and is independent of size and shape of gaussian surface.

16. In the situation, when gaussain surface is so chosen that there are some charges inside and some outside, the electric field (whose flux is calculated) is due to all the charges, both inside and outside the closed surface. However for electric flux, only inside charges contribute.

17. In case of a shpere (solid / hollow), electric field intensity at an external point varies as $\frac{1}{\mathrm{r}^{2}}$ where $\mathrm{r}$ is distance of the point from the centre of the sphere. In case of a line charge $\mathrm{E} \alpha \frac{1}{\mathrm{r}}$ and in case of an infinite plane sheet of charge, $\mathrm{E}$ does not depend upon $\mathrm{r}$.

18. Electrostatic force is a conservative force work done in bringing a test charge $\mathrm{q} _{0}$ from one point to another point is independent of path taken.

$$ \mathrm{W} _{\mathrm{A} \rightarrow \mathrm{B}}=\mathrm{q} _{0}\left[\mathrm{~V} _{\mathrm{B}}-\mathrm{V} _{\mathrm{A}}\right] $$

which is the difference of potential energy of test charge $\mathrm{q} _{0}$ between the final and initial position.

19. Electric potential at a point is numerically equal to the total amount of work done in bringing a unit positive charge from $\infty$ to that point ’ $r$ ‘.

$$ \mathrm{V}=\frac{\mathrm{w}}{\mathrm{q} _{0}}=-\int _{\infty}^{\mathrm{r}} \mathrm{Edr}=\frac{1}{4 \pi \varepsilon _{0}}=\frac{1}{4 \pi \varepsilon _{0}} \frac{\mathrm{q}}{\mathrm{r}} $$

For a point charge ’ $q$ ‘, electric potential exists at every point in an electric field regardless of whether there is a charge at that point or not.

20. For a charge configuration, the potential at a point $\mathrm{P}$ is given by

$$ \mathrm{V}=\frac{1}{4 \pi \varepsilon _{0}}\left(\frac{\mathrm{q} _{1}}{\mathrm{r} _{1 \mathrm{P}}}+\frac{\mathrm{q} _{2}}{\mathrm{r} _{2 \mathrm{P}}}+\ldots \ldots \frac{\mathrm{q} _{\mathrm{n}}}{\mathrm{r} _{\mathrm{nP}}}\right) $$

Where $r _{1 P}$ is the distance of $q$ from $P$, and so on.

21. Potential energy stored in a system of charges is the work done (by an external agency) in assembling the charges at their locations.

$$ \begin{aligned} & \mathrm{U}=\frac{1}{4 \pi \varepsilon _{0}}\left[\frac{\mathrm{q} _{1} \mathrm{q} _{2}}{\mathrm{r} _{12}}+\frac{\mathrm{q} _{1} \mathrm{q} _{3}}{\mathrm{r} _{13}}+\frac{\mathrm{q} _{2} \mathrm{q} _{3}}{\mathrm{r} _{23}}+\ldots\right] \\ & =\frac{1}{4 \pi \varepsilon _{0}} \frac{1}{2} \sum _{\mathrm{i}=1}^{\mathrm{n}} \sum _{\mathrm{j}=1}^{\mathrm{n}} \frac{\mathrm{q} _{1} \mathrm{q} _{\mathrm{j}}}{\mathrm{r} _{\mathrm{ij}}} \end{aligned} $$

22. Electric potential at a point due to an electric dipole with position vector ’ $r$ ’ is:

$ \mathrm{V}=\frac{1}{4 \pi \varepsilon _{0}} \frac{\overrightarrow{\mathrm{p}} \cdot \hat{\mathrm{r}}}{\mathrm{r}^{2}} $

On the dipole axis $\theta=0 _{0}$ or $\pi$

$ \mathrm{V}= \pm \frac{1}{4 \pi \varepsilon _{0}} \frac{\mathrm{p}}{\mathrm{r}^{2}} $

On the equitorial plane $\theta=\frac{\pi}{2}$

$ \mathrm{V}=0 $

23. Magnitude of electric field is given by change in the magnitude of potential per unit displacement (potential gradient) and its direction is always towards decreasing potential.

$$ \mathrm{E}=-\frac{\mathrm{dV}}{\mathrm{dr}} $$

24. Electrostatic field ’ $\mathrm{E}$ ’ is zero in the interior of conductor, just outside the surface of conductor, $\mathrm{E}$ is normal to the surface given by

$$ \mathbf{E}=\frac{\sigma}{\varepsilon _{0}} \hat{n} $$

Charges in a conductor can reside over the surface. Potential is constant within and on the surface of conductors. In a cavity, within a conductor, the electric field is zero. Any charge placed in the cavity, will appear over the surface of conductor.

25. For a parallel plate capacitor (with vacuum or air between the plates) its capacitance $\left(\mathrm{C}=\frac{\mathrm{Q}}{\mathrm{V}}\right)$ is defined as $\mathrm{C}=\frac{\mathrm{A} \varepsilon _{0}}{\mathrm{~d}}$ where $\mathrm{A}$ is the area of plate and $\mathrm{d}$ the separation between the plates. If the medium between the plates is filled with dielectric, its capacitance increases by $\mathrm{C}=\mathrm{kC} _{0}$, where $\mathrm{k}$ is the dielectric constant of the insulting substance.

26. For capacitors in the series combination, the total capacitance is given by

$$ \frac{1}{\mathrm{C} _{\mathrm{eq}}}=\frac{1}{\mathrm{C} _{1}}+\frac{1}{\mathrm{C} _{2}}+\ldots .+\frac{1}{\mathrm{Cn}} $$

where ’ $\mathrm{Q}$ ’ charge on each capacitor remain same. In parallel combination, the total $\mathrm{C}$ capacitance is given by

$$ \mathrm{C} _{\mathrm{eq}}=\mathrm{C} _{1}+\mathrm{C} _{2}+\ldots . .+\mathrm{C} _{\mathrm{n}} $$

where p.d. across the plates of each capacitors remains same.

27. The energy ’ $U$ ’ stored in a capacitor of capacitance $C _{1}$, with charge $Q$ and voltage $V$ is:

$$ \mathrm{U}=\frac{1}{2} \mathrm{QV}=\frac{1}{2} \mathrm{CV}^{2}=\frac{1}{2} \frac{\mathrm{Q}^{2}}{\mathrm{C}} $$

The energy density (energy per unit volume) in a region of $\mathbf{E}$ is $\frac{1}{2} \varepsilon _{0} \mathrm{E}^{2}$.

Average $\quad \quad \quad $ Coulomb Force/Electrostatic Force

1. In figure, two positive charges $q _{2}$ and $q _{3}$ fixed (symmetrically) along $y$-axis, exert a net electric force in the $+x$ direction. On charge $q _{1}$ fixed along the $x$-axis. If a charge ’ $+Q$ ’ is added at $(x, 0)$, the force on $\mathrm{q} _{1}$.

$~$

(1) shall increase along the positive $x$-axis

(2) shall decrease along the positive $x$-axis

(3) shall point along negative $x$-axis

(4) shall increas but direction charnges because of the intersection of $\mathrm{Q}$ with $\mathrm{q} _{2}$ and $\mathrm{q} _{3}\qquad\qquad$

Show Answer

Correct Answer: (1)

Solution:

As $\mathrm{q} _{2} ; \mathrm{q} _{3}$ are positive charges, and net force on $\mathrm{q} _{1}$ is along $+x$ direction, therefore $\mathrm{q} _{1}$ must be negative

When a positive charge ’ $\mathrm{Q}$ ’ is added $(x, 0)$. It will attract $-\mathrm{q} _{1}$ along $+x$ direction. Therefore force on $\mathrm{q} _{1}$ will increase along positive $x$ axis.

$~$

Correct answer: (1)

Easy $\quad \quad \quad $ Coulomb Force / Electrostatic Force

2. A small object with change $q$ and mass $m$ is attached to one end of a string of length ’ $L$ ‘. The other end is attached to a stationary support. The system is placed in a uniform horizontal electric field $\mathbf{E}$. In the presence of $\mathbf{E}$, the string makes a constant angle ’ $\theta$ ’ with the vertical. The sign and magnitude of $q$ is

(1) positive with magnitude $\frac{\mathrm{mg}}{\mathrm{E}}$

(2) positive with magnitude $\frac{\mathrm{mg}}{\mathrm{E}} \tan \theta$

(3) negative with magnitude $\frac{\mathrm{mg}}{\mathrm{E}}$

(4) positive with magnitude $\frac{\mathrm{mg}}{\mathrm{E}} \tan \theta$

Show Answer

Correct Answer: (2)

Solution:

For equilibrium $\mathrm{q}>0$,

$~$

$$ \begin{aligned} & \frac{ T \sin \theta}{ T \cos \theta}=\frac{\mathrm{qE}}{\mathrm{mg}} \\ & \Rightarrow \tan \theta=\frac{\mathrm{qE}}{\mathrm{mg}} \end{aligned} $$

$$ \text { or } q=\frac{m g \tan \theta}{E} $$

Correct answer: (2)

Average $\quad \quad \quad $ Coulomb Force/Electrostatic Force

3. A tiny spherical oil drop carrying a net charge $q$ is balanced in still air with a vertical uniform electric field strength $\frac{81 \pi}{7} \times \frac{10^{5} \mathrm{~V}}{\mathrm{~m}}$. When the field is switched off, the drop is observed to fall with terminal velocity $2 \times 10^{-3} \mathrm{~m} / \mathrm{s}$. Given $\mathrm{g}=9.8 \mathrm{~m} / \mathrm{s}^{2}$, viscosity of the air $=1.8 \times 10^{-5} \mathrm{Nsm}^{-2}$ and the density of oil $=\mathbf{9 0 0 0} \mathrm{kg} / \mathrm{m}^{3}$; the magnitude of $q$ is

(1) $1.6 \times 10^{-19} \mathrm{C}$

(2) $3.2 \times 10^{-19} \mathrm{C}$

(3) $4.8 \times 10^{-19} \mathrm{C} $

(4) $8.0 \times 10^{-19} \mathrm{C}$

Show Answer

Correct Answer: (4)

Solution:

When drop is balanced in still air under the effect of electric field then

$$ \begin{align*} & \mathrm{qE}=\frac{4}{3} \pi \mathrm{r}^{3} \rho \mathrm{g} \\ & \Rightarrow \mathrm{q}=\frac{4}{3 \mathrm{E}} \pi \mathrm{r}^{3} \rho \mathrm{g} \qquad\qquad \ldots . \ldots {(1)} \end{align*} $$

When electric field in switched off, drop falls with terminal velocity.

$$ \mathrm{V}=\frac{2 \mathrm{r}^{2}(\rho-\sigma) \mathrm{g}}{9 \eta} \text { or } \mathrm{r}=\left[\frac{9 \eta \mathrm{V}}{2(\rho-\sigma) \mathrm{g}}\right]^{1 / 2} \qquad\qquad \ldots . \ldots {(2)} $$

From (1) and (2)

$$ \Rightarrow \mathrm{q}=\frac{4}{3 \mathrm{E}} \pi \rho \mathrm{g}\left[\frac{9 \eta \mathrm{V}}{2(\rho-\sigma) g}\right]^{3 / 2} $$

or $\quad \mathrm{q}=\frac{4}{3} \times \frac{7}{81 \not{\pi}} \times \frac{\not{\pi}}{10^{5}} \times 900 \times 9.8\left[\frac{9 \times 1.8 \times 10^{-5} \times 2 \times 10^{-3}}{2 \times 900 \times 9.8}\right]^{3 / 2}$

or $\quad=8.0 \times 10^{-19} \mathrm{C}$

Correct answer: (4)

Difficult $\qquad \qquad \quad$ Coloumb Force/Electrostatics Force

4. Two identical charged sheres suspended from a common point by two massless strings of length $\ell$ are initially a distance $\mathrm{d}(\mathrm{d} < < \ell)$ apart because of their mutual repulsion. Now the charges begin to leak from both the spheres at a constant rate. As a result charges approach each other with a velocity ’ $V$ ‘. If ’ $x$ ’ is the distance between them at any instant, them

(1) $\quad \mathrm{V} \alpha x^{-1}$

(2) $\mathrm{V} \alpha x^{1 / 2}$

(3) $\mathrm{V} \alpha x$

(4) $\mathrm{V} \alpha x^{-1 / 2}$

Show Answer

Correct Answer: (4)

Solution:

$~$

$ \tan \theta=\frac{F}{m g} $

if $\tan \theta \simeq \theta$

$\Rightarrow \mathrm{F}=\operatorname{mg} \theta \simeq \frac{\operatorname{mg}(x / 2)}{\ell}$

$\simeq \frac{\operatorname{mg} x}{2 \ell}$

or $\quad \frac{\operatorname{mgx}}{2 \ell}=\frac{\mathrm{KQ}^{2}}{x^{2}}$

$\Rightarrow \mathrm{Q}^{2}=\left(\frac{\mathrm{mg}}{2 \mathrm{k} \ell}\right) x^{3}$

or $\quad \mathrm{Q}^{2} \alpha x^{3}$

$ \mathrm{Q} \alpha x^{3 / 2}$

$ \Rightarrow \frac{\mathrm{dQ}}{\mathrm{dt}}=\mathrm{C} \times \frac{3}{2} x^{1 / 2} \frac{\mathrm{d} x}{\mathrm{dt}} $

$ \text { or } \frac{\mathrm{d} x}{\mathrm{dt}}=\frac{2 x^{-1 / 2}}{3 \mathrm{C}} \frac{\mathrm{dQ}}{\mathrm{dt}}$

$ \Rightarrow \frac{\mathrm{d} x}{\mathrm{dt}} \alpha x^{-1 / 2}$

$ \Rightarrow \mathrm{v} \alpha x^{-1 / 2}$

Correct answer: (4)

Average $\qquad \qquad \quad$ Coloumb Force/Electrostatics Force

5. Three charges $\frac{\mathrm{q}}{3}, \frac{\mathrm{q}}{3}$ and $\frac{\mathrm{q}}{3}$ are placed at the circuference of a circle of radius ’ $R$ ’ at points $A$, $B$, and $C$ shown in the diagram. The magnitude of force between charges placed at $C$ and $B$ is

$~$

(1) $\frac{\mathrm{kq}^{2}}{36 \mathrm{R}^{2}}$

(2) $\frac{\mathrm{kq}^{2}}{9 \mathrm{R}^{2}}$

(3) $\frac{\mathrm{kq}^{2}}{27 \mathrm{R}^{2}}$

(4) $\frac{4 \mathrm{kq}^{2}}{9 \mathrm{R}^{2}}$

Show Answer

Correct Answer: (3)

Solution:

as $\sin 60^{\circ}=\frac{\mathrm{BC}}{\mathrm{AB}}$

$\Rightarrow \mathrm{BC}=\mathrm{AB} \sin 60^{\circ}$

$ =\not 2 \mathrm{R} \frac{\sqrt{3}}{\not 2}=\mathrm{R} \sqrt{3} $

$\mathrm{F} _{\mathrm{CB}}=\frac{\mathrm{k} \frac{\mathrm{q}}{3} \frac{\mathrm{q}}{3}}{(\mathrm{R} \sqrt{3})^{2}}=\frac{\mathrm{kq}^{2}}{27 \mathrm{R}^{2}}$

Correct answer: (3)

Easy $\quad \quad \quad $ Electric Field

6. Two point charges $Q$ and $-2 Q$ are placed some distance apart. If the electric field at the location of $Q$ is $E$, then the electric field at the location of $-2 Q$ will be

(1) $-\mathrm{E} / 2$

(2) $-\mathrm{E} \quad \quad\quad $

(3) $-3 \mathrm{E} / 2$

(4) $-2 \mathrm{E}$

Show Answer

Correct Answer: (1)

Solution:

Let ’ $r$ ’ be the distance between $Q$ and $-2 Q$. Electric field due to $-2 Q$ at the location of $Q$ is

$$ \begin{equation*} \mathrm{E}=\frac{\mathrm{k}(-2 \mathrm{Q})}{\mathrm{r}^{2}} \qquad\qquad \ldots . \ldots {1} \end{equation*} $$

Electric field due to $+\mathrm{Q}$ at the location of $-2 \mathrm{Q}$ is

$$ \begin{equation*} E^{\prime}=\frac{k(Q)}{r^{2}} \qquad\qquad \ldots . \ldots {2} \end{equation*} $$

From (1) & (2)

$$ \begin{aligned} & \frac{E^{\prime}}{E}=-\frac{1}{2} \\ & E^{\prime}=-\frac{1}{2} E \end{aligned} $$

Correct answer: (1)

Difficult $\quad \quad \quad $ Electric Field

7. Let there be a spherical symmetric charge distribution with charge density varying as $\rho(x)=\rho _{0}\left(\frac{5}{4}-\frac{r}{R}\right)$ upto $r=k$ and $\rho(r)=0$ for $r>R$, where $r$ is the distance from the origin. The elctric field at a distance $r(r<R)$ from the origin is given by

(1) $\frac{\rho \mathrm{r}}{4 \varepsilon _{0}}\left(\frac{5}{3}-\frac{\mathrm{r}}{\mathrm{R}}\right)$

(2) $\frac{4 \rho _{0} \mathrm{r}}{3 \varepsilon _{0}}\left(\frac{5}{3}-\frac{\mathrm{r}}{\mathrm{R}}\right)$

(3) $\frac{\rho _{0} \mathrm{r}}{3 \varepsilon _{0}}\left(\frac{5}{3}-\frac{\mathrm{r}}{\mathrm{R}}\right)$

(4) $\frac{4 \pi \rho _{0} \mathrm{r}}{3 \varepsilon _{0}}\left(\frac{5}{3}-\frac{\mathrm{r}}{\mathrm{R}}\right)$

Show Answer

Correct Answer: (4)

Solution:

$~$

$$ \begin{aligned} & \mathrm{q}=\int \rho \mathrm{dv}=\int _{0}^{\mathrm{r}} \rho _{0}\left(\frac{5}{4}-\frac{x}{\mathrm{R}}\right) 4 \pi x^{2} \mathrm{~d} x \\ & =\rho _{0}\left[\frac{5}{4} 4 \pi \frac{x^{3}}{3}-\frac{4 \pi}{\mathrm{R}} \frac{x^{4}}{4}\right] _{0}^{\mathrm{r}} \\ & =\rho _{0}\left[\frac{5}{4} \pi \mathrm{r}^{3}-\frac{\pi \mathrm{r}^{4}}{\mathrm{R}}\right] \qquad\qquad \ldots . \ldots (1) \end{aligned} $$

Electric field intensity at a point on this sheprical surface is

$$ \begin{gather*} \mathrm{E}=\frac{1}{4 \pi \varepsilon _{0}} \frac{\mathrm{q}}{\mathrm{r}^{2}}=\frac{1}{4 \pi \varepsilon _{0}} \frac{\rho _{0}}{\mathrm{r}^{2}}\left[\frac{5}{3} \pi \mathrm{r}^{3}-\frac{\pi \mathrm{r}^{4}}{\mathrm{R}}\right] \qquad \qquad \mathrm{From} (1) \\ \\ \Rightarrow \mathrm{E}=\frac{\rho _{0} \mathrm{r}}{4 \varepsilon _{0}}\left[\frac{5}{3}-\frac{\mathrm{r}}{\mathrm{R}}\right] \end{gather*} $$

Correct answer: (4)

Difficult $\quad \quad \quad $ Electric Field

8. A positively charged metal ring a radius $\mathrm{R}$ is fixed in the $x-y$ plane with its centre at origin ’ $O$ ‘. A negatively charged particle $P$ is released from rest at the point $\left(0,0, z _{0}\right)$ where $z _{0}>0$. Then the motion of charged particle $P$ is

(1) periodic provided $\mathrm{z} _{0}<\mathrm{R}$

(2) simple harmonic for all the values of $z _{0}$

(3) approximately simple harmonic provided $\mathrm{z} _{0} < < \mathrm{R}$

(4) such that it crosses centre of ring ’ $\mathrm{O}$ ’ and continues to move along $-\mathrm{Z}$ axis

Show Answer

Correct Answer: (3)

Solution:

The elctric field at the point ’ $\mathrm{P}$ ’ due to uniformly charged metal ring of total charge ’ $Q$ ’ is given by:

$$ \mathrm{E}=\frac{1}{4 \pi \varepsilon _{0}} \frac{\mathrm{Qz} _{0}}{\left(\mathrm{R}^{2}+\mathrm{z} _{0}^{2}\right)^{3 / 2}} $$

Away from the centre of the ring.

Now force on ’ $-q$ ’ at $P$ will be-

$$ \begin{aligned} & \mathrm{F}=-\mathrm{qE} \\ & =-\frac{1}{4 \pi \varepsilon _{0}} \frac{\mathrm{Qqz} _{0}}{\left(\mathrm{R}^{2}+\mathrm{z} _{0}^{2}\right)^{3 / 2}} \end{aligned} $$

$~$

[Acts towards centre ’ $\mathrm{O}$ ’ of the ring.]

For $\mathrm{z} _{0} < < \mathrm{R}$

$$ \begin{aligned} \mathrm{F} & =-\frac{1}{4 \pi \varepsilon _{0}} \frac{\mathrm{Qqz} _{0}}{\mathrm{R}^{3}} \\ \text { or } \quad \mathrm{F} & =-\left(\frac{1}{4 \pi \varepsilon _{0}} \frac{\mathrm{Qq}}{\mathrm{R}^{3}}\right) \mathrm{z} _{0} \\ \text { or } \quad \mathrm{F} & =-K \mathrm{z} _{0} \end{aligned} $$

Where $\mathrm{K}=\frac{1}{4 \pi \varepsilon _{0}} \frac{\mathrm{Qq}}{\mathrm{R}^{3}}=$ Constant

The above equation represents S.H.M. as $\mathrm{F} \alpha \mathrm{z} _{0}$ and directed towards centre.

Correct answer: (3)

Average $\quad \quad \quad $ Electric Field

9. Three infinitely charged sheets are kept parallel to $x y$ plane having charge densities as shown. The electric field strength at the point ’ $P$ ’ is:

$~$

(1) $\frac{-4 \sigma}{\varepsilon _{0}} \hat{\mathrm{k}}$

(2) $\frac{4 \sigma}{\varepsilon _{0}} \hat{\mathrm{k}}$

(3) $\frac{-2 \sigma}{\varepsilon _{0}} \hat{\mathrm{k}}$

(4) $\frac{2 \sigma}{\varepsilon _{0}} \hat{\mathrm{k}}$

Show Answer

Correct Answer: (3)

Solution:

For infinite long sheet, electric field is given by $\mathbf{E}=\frac{\sigma}{2 \varepsilon _{0}} \hat{\mathrm{r}}$ (Independent of r). Now electric field due to charge density $+\sigma$, at $\mathrm{P}$, is

$$ \mathbf{E} _{1}=\frac{\sigma}{2 \varepsilon _{0}}(-\hat{\mathrm{k}}) $$

Electric field due to charge density $-2 \sigma$ at $\mathrm{P}$, is

$$ \mathbf{E} _{2}=\frac{2 \sigma}{2 \varepsilon _{0}}(-\hat{\mathrm{k}}) $$

Electric field due to change density $-\sigma$ at point $\mathrm{P}$, is

$$ \mathbf{E} _{3}=\frac{\sigma}{2 \varepsilon _{0}}(-\hat{\mathrm{k}}) $$

Hence total electric field at point ’ $\mathrm{P}$ ’ is $\mathbf{E}=\mathbf{E} _{1}+\mathbf{E} _{2}+\mathbf{E} _{3}$

or $\mathrm{E} _{\text {net }}=\frac{\sigma}{2 \varepsilon _{0}}(-\hat{\mathrm{k}})+\frac{2 \sigma}{2 \varepsilon _{0}}(-\hat{\mathrm{k}})+\frac{\sigma}{2 \varepsilon _{0}}(-\hat{\mathrm{k}})$

or $\quad \mathrm{E} _{\text {net }}=\frac{2 \sigma}{\varepsilon _{0}}(-\hat{\mathrm{k}})=-\frac{-\sigma}{\varepsilon _{0}} \hat{\mathrm{k}}$

Correct answer: (3)

Average $ \qquad \quad \quad $ Continuous Charge Distribution

10. A thin semi-circular ring of radius $r$ has a positive charge $q$ distributed uniformly over it, as shown in the figure. The net electric field at the centre ’ $O$ ’ is

$~$

(1) $-\frac{1}{4 \pi^{2} \varepsilon _{0}} \frac{\mathrm{q}}{\mathrm{r}^{2}} \hat{\mathrm{j}}$

(2) $-\frac{1}{2 \pi^{2} \varepsilon _{0}} \frac{\mathrm{q}}{\mathrm{r}^{2}} \hat{\mathrm{j}}$

(3) $\frac{q}{2 \pi^{2} \varepsilon _{0} r^{2}} \hat{j}$

(4) $\frac{\mathrm{q}}{4 \pi^{2} \varepsilon _{0} \mathrm{r}^{2}} \hat{\mathrm{j}}$

Show Answer

Correct Answer: (3)

Solution:

Charge on element ’ $\mathrm{d} \ell$ ‘, $\mathrm{dq}=\lambda \mathrm{d} \ell=\lambda \mathrm{rd} \theta$

Electric field due to this element is $\mathrm{dE}=\frac{1}{4 \pi \varepsilon _{0}} \frac{\lambda \mathrm{d} \theta}{\mathrm{r}^{2}}$

As shown in the figure, one can realise that components along $x-$ direction cancel and in pairs. Hence total electric field.

$~$

$~$

$\mathrm{E}=\int _{0}^{180^{\circ}} \mathrm{dE} \sin \theta(-\mathrm{j}) \quad[\operatorname{along}(-\mathrm{j})]$

$~$

or $\quad E=\frac{1}{4 \pi \varepsilon_0} \frac{\lambda^{180^{\circ}}}{\mathrm{r}} \int_0^{\mathrm{dE}} \sin \theta(-\mathrm{j})=\frac{\mathrm{q}}{2 \pi^2 \varepsilon_0 \mathrm{r}^2} \hat{\mathrm{j}}$

Correct answer: (3)

Average $ \quad \quad \quad$ Electric Potential

11. In the figure given below, equipotential lines are drawn at $0,20 \mathrm{~V}$ and $40 \mathrm{~V}$.

$~$

The work done in moving a change of $+3 \mu \mathrm{C}$ from position $A$ to position $\mathrm{B}$ is-

(1) $4 \mu \mathrm{J}$

(2) $8 \mu \mathrm{J}$

(3) $12 \mu \mathrm{J}$

(4) $120 \mu \mathrm{J}$

Show Answer

Correct Answer: (4)

Solution:

Work done is independent of path taken. Hence

$$ \begin{aligned} \mathrm{w} _{\mathrm{A} \rightarrow \mathrm{B}} & =\Delta \mathrm{q}\left(\mathrm{V} _{\mathrm{B}}-\mathrm{V} _{\mathrm{A}}\right) \\ & =3 \mu \mathrm{C}(40 \mathrm{~V}-0 \mathrm{~V}) \\ & =120 \mu \mathrm{J} \end{aligned} $$

Correct answer: (4)

Average $\quad \quad \quad $ Electric Potential

12. The infinite long wire having linear change density ’ $\lambda$ ’ is lying in $x y$ plane. A point charge $q _{0}$ is moved from point A to point $B$ through the circular path $A B$ of radius ’ $a$ ’ centred at $C$ in the same plane as shown in the figure. The amount of work done is equal to:

$~$

(1) $\frac{\mathrm{q} _{0}}{2 \pi \varepsilon _{0}} \ell n \frac{2}{3}$

(2) $\frac{\mathrm{q} _{0} \lambda}{2 \pi \varepsilon _{0}} \ell n \frac{3}{2}$

(3) $\frac{\mathrm{q} _{0} \lambda}{2 \pi \varepsilon _{0}} \ell n \frac{2}{3}$

(4) $\frac{\mathrm{q} _{0} \lambda}{\sqrt{2} \pi \varepsilon _{0}}$

Show Answer

Correct Answer: (2)

Solution:

Electric field due to infinite long wire is equal to $E=\frac{\lambda}{2 \pi \varepsilon _{0} \mathrm{r}}$

as $\mathrm{V}=-\int \mathrm{Edr} \Rightarrow \mathrm{V}=-\frac{\lambda}{2 \pi \varepsilon _{0}} \ell n (\mathrm{r})$

So potentialat A, $\mathrm{V} _{\mathrm{A}}=\frac{-\lambda}{2 \pi \varepsilon _{0}} \ell n (3 \mathrm{a})$

and potential at $\mathrm{B}, \mathrm{V} _{\mathrm{B}}=\frac{-\lambda}{2 \pi \varepsilon _{0}} \ell n (2 \mathrm{a})$

Work done in moving $\mathrm{q} _{0}$ from A to $\mathrm{B}$

$$ \begin{aligned} & \mathrm{w} _{\mathrm{A} \rightarrow \mathrm{B}}=\mathrm{q} _{0}\left(\mathrm{~V} _{\mathrm{B}}-\mathrm{V} _{\mathrm{A}}\right) \\ & =\mathrm{q} _{0}\left[\left(-\frac{\lambda}{2 \pi \varepsilon _{0}} \ell n (2 \mathrm{a})\right)-\left(-\frac{\lambda}{2 \pi \varepsilon _{0}} \ell n (3 \mathrm{a})\right)\right] \\ & =\frac{\lambda}{2 \pi \varepsilon _{0}} \mathrm{q} _{0} \ell n \left(\frac{3}{2}\right) \end{aligned} $$

Correct answer is (2)

Difficult $\quad \quad \quad $ Electric Potential

13. A solid conducting sphere having a charge $Q$ is surrounded by an uncharged co-centric conducting hollow spherical shell. The p.d. between the surface of solid sphere and outer surface of hollow shell is ’ $V$ ‘. Now if outer shell is given a charge of $-3 Q$, the new p.d between the same two surfaces is

(1) $-2 \mathrm{~V}$

(2) $\mathrm{V}$

(3) $2 \mathrm{~V}$

(4) $4 \mathrm{~V}$

Show Answer

Correct Answer: (2)

Solution:

Initially, $\mathrm{V}=\frac{1}{4 \pi \varepsilon _{0}} \frac{\mathrm{Q}}{\mathrm{r} _{1}}-\frac{1}{4 \pi \varepsilon _{0}} \frac{\mathrm{Q}}{\mathrm{r} _{2}}$

When outer shell is given a charge of ’ $-3 Q$ '

$~$

$ \mathrm{V}^{\prime}=\frac{1}{4 \pi \varepsilon _{0}}\left[\left(\frac{\mathrm{Q}}{\mathrm{r} _{1}}-\frac{3 \mathrm{Q}}{\mathrm{r} _{2}}\right)-\left(\frac{\mathrm{Q}}{\mathrm{r} _{2}}-\frac{3 \mathrm{Q}}{\mathrm{r} _{2}}\right)\right] $

or $\quad \mathrm{V}^{\prime}=\frac{1}{4 \pi \varepsilon _{0}}\left[\begin{array}{ll}\frac{\mathrm{Q}}{\mathrm{r} _{1}} & -\frac{\mathrm{Q}}{\mathrm{r} _{2}}\end{array}\right]=\mathrm{V}$

Correct answer: (2)

Average $\quad \quad \quad $ Electric Potential

14. A conducting spherical bubble of radius $a$ and thickness $t(t < < a)$ is charged to a potential $V$. Now it collapses to form a spherical droplet. The potential of the droplet is

(1) $\mathrm{V}$

(2) $V\left[\frac{a}{t}\right]^{1 / 3}$

(3) $V\left[\frac{a}{3 t}\right]^{1 / 3}$

(4) $\quad V\left[3 a^{2} t\right]^{1 / 3}$

Show Answer

Correct Answer: (3)

Solution:

As volume remains constant

$$ \begin{equation*} \frac{4}{3} \pi \mathrm{R}^{3}=4 \pi \mathrm{a}^{2} \mathrm{t} \qquad\qquad \ldots . \ldots {(1)} \end{equation*} $$

Where $\mathrm{R}$ is the radius of droplet formed. Using conservation of charge. $[\mathrm{Q}=\mathrm{CV}]$

$$ \begin{align*} & \left(4 \pi \varepsilon _{0} \mathrm{R}\right) \mathrm{V}^{\prime}=\left(4 \pi \varepsilon _{0} \mathrm{a}\right) \mathrm{V} \qquad\qquad \ldots . \ldots {(2)}\\ & \Rightarrow \mathrm{V}^{\prime}=\frac{\mathrm{aV}}{\mathrm{R}}=\frac{\mathrm{aV}}{\left(3 \mathrm{a}^{2} \mathrm{t}\right)^{1 / 3}}=\mathrm{V}\left[\frac{\mathrm{a}}{3 \mathrm{t}}\right]^{1 / 3} \qquad \qquad [ \text{Using} (1)] \end{align*} $$

Correct answer is (3)

Difficult $\qquad \quad \quad$ Electrostatic Potential Energy

15. Two point charges each of charge $q$ are fixed at $(+a, 0)$ and $(-a, 0)$. Another positive charge $q$ placed at the origin is free to move along $x$-axis. The charge $q$ at origin in equilibrium will have:

(1) maximum force and minimum potential energy

(2) minimum force and maximum potential energy

(3) maximum force and maximum potential energy

(4) minimum force and minimum potential energy

Show Answer

Correct Answer: (4)

Solution:

Net force on $q$ at origin is

$$ \mathbf{F}=\mathbf{F} _{1}+\mathbf{F} _{2}=\frac{1}{4 \pi \varepsilon _{0}} \frac{\mathrm{q}^{2}}{\mathrm{a}^{2}} \hat{\mathrm{i}}+\frac{1}{4 \pi \varepsilon _{0}} \frac{\mathrm{q}^{2}}{\mathrm{a}^{2}}(-\hat{\mathrm{i}})=0 $$

Let the charge $\mathrm{q}$ is displaced from origin to position $(x, 0)$. At this point net force on the charge ’ $q$ ’ due to two end charges is

$~$ $~$

$\mathrm{F}=\frac{1}{4 \pi \varepsilon _{0}} \mathrm{q}^{2}\left[\frac{1}{(\mathrm{a}-x)^{2}}+\frac{1}{(\mathrm{a}+x)^{2}}\right]$

as $\quad|\mathrm{F}|=\frac{\mathrm{dU}}{\mathrm{d} x}$ for conservative fields.

$\Rightarrow \frac{\mathrm{dU}}{\mathrm{d} x}=\frac{1}{4 \pi \varepsilon _{0}} \mathrm{q}^{2}\left[\frac{1}{(\mathrm{a}-x)^{2}}+\frac{1}{(\mathrm{a}+x)^{2}}\right]$

For $\mathrm{U}$ to be minimum $\frac{\mathrm{d}^{2} \mathrm{U}}{\mathrm{d} x^{2}}>\mathrm{O}$.

$\Rightarrow(\mathrm{a}-x)^{2}=(\mathrm{a}+x)^{2} \Rightarrow x=0$

Hence charge at origin will have minimum minimum potential energy.

Correct answer: (4)

Average $\qquad \quad \quad$ Electrostatic Potential Energy

16. A proton and an $\alpha$-particle are situated at a distance ’ $r$ ’ apart. At very large distance apart, when released, the kinetic energy of proton will be

(1) $\frac{2 \mathrm{Ke}^{2}}{\mathrm{r}}$

(2) $\frac{8 \mathrm{Ke}^{2}}{5 \mathrm{r}}$

(3) $\frac{\mathrm{Ke}^{2}}{\mathrm{r}}$

(4) $\frac{8 \mathrm{Ke}^{2}}{\mathrm{r}}$

Show Answer

Correct Answer: (2)

Solution:

According to conservation of energy-

$$ \begin{gather*} \frac{\mathrm{K}(\mathrm{e})(2 \mathrm{e})}{\mathrm{r}}=(\mathrm{K} . \mathrm{E}) _{\mathrm{P}}+(\text { K.E. }) _{\alpha} \\ =\frac{1}{2} \mathrm{~m} _{\mathrm{p}} \mathrm{v} _{\mathrm{p}}^{2}+\frac{1}{2} \mathrm{~m} _{\alpha} \mathrm{v} _{\alpha}^{2} \qquad \qquad \ldots . \ldots {(1)} \end{gather*} $$

According to conservation of momentum.

$$ \begin{equation*} 0=\mathrm{m} _{\mathrm{p}} \mathrm{v} _{\mathrm{p}}=\mathrm{m} _{\alpha} \mathrm{v} _{\alpha} \qquad\qquad \ldots . \ldots {(2)} \end{equation*} $$

$$ \begin{align*} & \mathrm{m} _{\mathrm{p}} \mathrm{v} _{\mathrm{p}}=\mathrm{m} _{\alpha} \mathrm{v} _{\alpha} \\ & 1 \times \mathrm{v} _{\mathrm{p}}=4 \times \mathrm{v} _{\alpha} \\ & \mathrm{v} _{\alpha}=\frac{\mathrm{v} _{\mathrm{p}}}{4} \qquad \qquad \ldots . \ldots {(3)} \end{align*} $$

From(1) and (3)

(K.E) $\mathrm{P}=\frac{1}{2} \mathrm{~m} _{\mathrm{p}} \mathrm{v} _{\mathrm{p}}^{2}=\frac{8}{5} \frac{\mathrm{Ke}^{2}}{\mathrm{r}}$

Correct answer: (2)

Average $\qquad \quad \quad$ Electric Flux

17. The electric field strength in a region is given as $\mathbf{E}=\frac{x \hat{i}+y \hat{j}}{x^{2}+y^{2}}$. The net charge inside a sphere of radius ’ $a$ ’ with its centre at origin will be

$~$ $~$

(1) zero

(2) $4 \pi \mathrm{a} \varepsilon _{0}$

(3) $2 \pi \mathrm{a} \varepsilon _{0}$

(4) $8 \pi \mathrm{a} \varepsilon _{0}$

Show Answer

Correct Answer: (2)

Solution:

At a point $(x, y, z)$ on the sphere a unit vector perpendicular to surface can be written as $\hat{n}=\frac{x}{a} \hat{i}+\frac{y}{a} \hat{j}+\frac{z}{a} \hat{k}$ and as given that $\mathbf{E}=\frac{x \hat{\mathrm{i}}+\mathrm{y} \hat{\mathrm{j}}}{x^{2}+\mathrm{y}^{2}}$

$\Rightarrow \mathbf{E} \hat{n}=\left(\frac{x i+y}{x^{2}+y^{2}}\right) \cdot\left(\frac{x}{a} \hat{i}+\frac{y}{a} \hat{j}+\frac{z}{a} \hat{k}\right)$

$=\frac{1}{\mathrm{a}}$, which is independent of co-ordinate of point ’ $\mathrm{P}$ ’ lying on sphere. Also $\mathbf{E} . \hat{\mathrm{n}}=$ total flux passing through the unit area of sphere.

$\Rightarrow$ Total flux passing through sphere.

$$ \begin{aligned} & \phi=(\mathbf{E} \cdot \hat{\mathrm{n}}) 4 \pi \mathrm{a}^{2}=\frac{\mathrm{q}}{\varepsilon _{0}} \\ & \Rightarrow \mathrm{q}=\varepsilon _{0} \frac{1}{\mathrm{a}} \times 4 \pi \mathrm{a}^{2}=4 \pi \mathrm{a} \varepsilon _{0} \end{aligned} $$

Correct answer: (2)

Average $\qquad \quad \quad$ Electric Flux

18. Consider an electric field, $\mathbf{E}=\mathrm{E} _{0} \hat{x}$, where $\mathrm{E} _{0}$ is a constant. The flux through the shaded region (as shown in the figure) due to this field is

$~$

(1) $2 \mathrm{E} _{0} \mathrm{a}^{2}$

(2) $\sqrt{2} \mathrm{E} _{0} \mathrm{a}^{2}$

(3) $\mathrm{E} _{0} \mathrm{a}^{2}$

(4) $\frac{E _{0} a^{2}}{\sqrt{2}}$

Show Answer

Correct Answer: (3)

Solution:

The shaded area $\mathbf{A}=\mathbf{P} \boldsymbol{\theta} \times \mathbf{P S}$

$\therefore \mathbf{A}=\left(\mathrm{a}^{2} \hat{\mathrm{i}}-\mathrm{a}^{2} \hat{\mathrm{k}}\right)$

Electric flux $\phi=\mathbf{E} . \mathbf{A}$

$$ \begin{aligned} & \phi=\left(\mathrm{E} _{0} \hat{\mathrm{i}}\right) \mathrm{a}^{2} \hat{\mathrm{i}}-\mathrm{a}^{2} \hat{\mathrm{k}} \\ & \phi=\mathrm{E} _{0} \mathrm{a}^{2} \end{aligned} $$

Correct answer: (3)

Average $\qquad \quad \quad$ Electric Flux

19. Three charges $q _{1}=1 \mu C, q _{2}=2 \mu C$, and $q _{3}=-3 \mu C$ and four surfaces $S _{1}, S _{2}, S _{3}$ and $S _{4}$ are shown in figure. The flux emerging through surface $S _{2}$ in $\mathrm{Nm}^{2} / \mathrm{C}$ is

$~$

(1) $36 \pi \times 10^{3}$

(2) $-36 \pi \times 10^{3}$

(3) $36 \pi \times 10^{9}$

(4) $-36 \pi \times 10^{9}$

Show Answer

Correct Answer: (2)

Solution:

As the flux is due to all the charges present inside the gaussian surface. hence flux through $\mathrm{S} _{2}$,

$\phi _{2}=\frac{\text { Total charge enclosed }}{\varepsilon _{0}}$

or $\quad \phi _{2}=\frac{(2 \mu \mathrm{C}-3 \mu \mathrm{C})}{\varepsilon _{0}}=-1 \times 10^{-6} \times 4 \pi \mathrm{k}$

or $\phi _{2}=-1 \times 10^{-6} \times 4 \pi \times 9 \times 10^{9}$

or $\phi _{2}=-36 \pi \times 10^{3} \mathrm{Nm}^{2} / \mathrm{C}$

Correct answer: (2)

Difficult $\qquad \quad \quad$ Electric Flux

20. The volume charge density as a function of distance $x$ from one face inside a unit cube is varrying as shown in the figure. The total flux through the cube is (if $\rho _{0}=8.85 \times 10^{-12} \mathrm{C} / \mathrm{m}^{3}$ ).

$~$

(1) $\frac{1}{4}$ SIUnits

(2) $\frac{1}{2}$ SIUnits

(3) $\frac{3}{4}$ SIUnits

(4) 1 SI Units

Show Answer

Correct Answer: (3)

Solution:

Let ’ $q$ ’ is the total charge enclosed within the cube.

$\mathrm{q}=$ Area of curve (trapezium)

$$ =\frac{3}{4} \rho _{0} $$

As flux passing throught the cube will be given by

$$ \begin{aligned} \phi=\frac{\mathrm{q}}{\varepsilon _{0}} & =\frac{\frac{3}{4} \rho _{0}}{\varepsilon _{0}}=\frac{\frac{3}{4} \times 8.85 \times 10^{-12}}{8.85 \times 10^{-12}} \\ \phi & =\frac{3}{4} \text { SI Units } \end{aligned} $$

Correct answer: (3)

Average $\qquad \quad \quad$ Conductors in Electrostatics Equilibrium

21. A cavity is created in the conducting shell having uniform charge distribution of +8 units as shown in the figure. Now if a charge of +4 units in placed at the centre of cavity. Which set of equations correctly describe the magnitude of electric field.

$~$

	$r < r_1$	$r_1< r< r_2$	$ \quad r < r2$
(1)	$\mathrm{E}=0$	$\mathrm{E}=\frac{K 12 q}{r^{2}}$	$mathrm{E}=\frac{K 8 q}{r^{2}}$
(2)	$\mathrm{E}=\frac{K 4 q}{r^{2}}$	$\mathrm{E}=0$	$\mathrm{E}=\frac{K 12 q}{r^{2}}$
(3)	$\mathrm{E}=\frac{K 4 q}{r^{2}}$	$\mathrm{E}=\frac{K 8 q}{r^{2}}$	$\mathrm{E}=0$
(4)	$\mathrm{E}=\frac{K 4 q}{r^{2}}$	$\mathrm{E}=0$	$\mathrm{E}=\frac{K 8 q}{r^{2}}$

Show Answer

Correct Answer: (2)

Solution:

$E _{\text {inside }}$ cavity will be non-zero having value $E=\frac{K 4 q}{r^{2}}\left(\right.$ for $\left.r<r _{1}\right)$.

In electrostatic equilibrium, field inside conductor is zero (for $\mathrm{r} _{1}<\mathrm{r}<\mathrm{r} _{2}$ ).

As charge placed at cavity, will appear over the surface of conductor, hance $\mathrm{q}=8+4=12$ units.

Thus, $E=\frac{K 12 q}{r^{2}}\left(\right.$ for $\left.r>r _{2}\right)$

Correct answer: (2)

Average $\qquad \quad \quad$ Conductors in Electrostatics Equilibrium

22. A spherical shell with an inner surface of radius ’ $a$ ’ and an outer surface of radius ’ $b$ ’ is made of conducting material. A point charge ’ $+Q$ ’ is placed at the centre of the spherical shell and a total charge $-q$ is placed on the shell.

$~$

Assuming that the electrostatic potential is zero at infinite distance. The elctrostatic potential at a distance ’ $R$ ’ from the conducting shell where $b>R>a$ is

(1) $\frac{\mathrm{KQ}}{\mathrm{a}}$

(2) $\frac{\mathrm{KQ}}{\mathrm{R}}$

(3) $\frac{\mathrm{K}(\mathrm{Q}-\mathrm{q})}{\mathrm{R}}$

(4) $\frac{\mathrm{K}(\mathrm{Q}-\mathrm{q})}{\mathrm{b}}$

Show Answer

Correct Answer: (4)

Solution:

The charge ’ $Q$ ’ put at the centre will appear on the surface of conducting shell. So that field inside the shell remains zero in equilibrium. Also electrostatic potential in constant throughout the volume and surface of the conductor.

Correct answer: (4)

Average $\qquad \quad \quad $ Dielectrics

23. A parallel plate capacitor with air as the dielectric and having the same thickness as the separation between the plates is introduced so as to fill one fourth of the capacitor as shown in the figure.

$\mathrm{K}$	air
$\mathrm{A} / 4$	$3 \mathrm{~A} / 4$

The new capacitance will be-

(1) $(\mathrm{K}+3) \frac{\mathrm{C}}{4}$

(2) $(K+2) \frac{C}{4}$

(3) $(\mathrm{K}+1) \frac{\mathrm{C}}{4}$

(4) $\frac{\mathrm{KC}}{4}$

Show Answer

Correct Answer: (1)

Solution:

The capacitance $\mathrm{C}=\frac{\mathrm{A} \varepsilon _{0}}{\mathrm{~d}}$ (without dielectric).

Let $C _{1}$ be the capacitance of dielectric part then $C _{1}=\frac{K \varepsilon _{0} A / 4}{d}$ and $C _{2}$ (without dielectric part) $=\frac{\varepsilon _{0} 3 \mathrm{~A} / 4}{\mathrm{~d}}$

The combination acts like parallel combination such that:

$$ \begin{aligned} & \mathrm{C} _{\text {net }}=\mathrm{C} _{1}+\mathrm{C} _{2}=\frac{\mathrm{K} \varepsilon _{0} \mathrm{~A} / 4}{\mathrm{~d}}+\frac{\frac{3}{4} \varepsilon _{0} \mathrm{~A}}{\mathrm{~d}} \\ & \mathrm{C} _{\text {net }}=(\mathrm{K}+3) \frac{\mathrm{C}}{4} \end{aligned} $$

Correct answer: (1)

Average $\qquad \quad \quad$ Dielectrics

24. Two identical capacitors 1 and 2 are connected in series to a battary of ’ $V$ ’ volts as shown in the figure. Capacitor 2 contains a dielectric slab of dielectric constant $K$ and $Q _{1}, Q _{2}$ are the charges stored in the capacitors repectively. Now the dielectric slab is removed and the corresponding charges are $Q _{1}^{\prime}$ and $Q _{2}^{\prime}$. Then the ratio $\frac{\mathrm{Q} _{2}^{\prime}}{\mathrm{Q} _{2}^{\prime}}$ is

$~$

(1) $\frac{\mathrm{K}+1}{\mathrm{~K}}$

(2) $\frac{\mathrm{K}+1}{2}$

(3) $\frac{\mathrm{K}+1}{2 \mathrm{~K}}$

(4) $\frac{K}{2}$

Show Answer

Correct Answer: (3)

Solution:

Let ’ $\mathrm{C} _{0}$ ’ is the capacitance of each capacitor without dielectric.

$$ \Rightarrow \mathrm{C} _{1}=\mathrm{C} _{0} \text { and } \mathrm{C} _{2}=\mathrm{kC} _{0} $$

Before dielectric is removed.

$$ \begin{align*} & \mathrm{C} _{\text {net }}=\frac{\mathrm{C} _{1} \mathrm{C} _{2}}{\mathrm{C} _{1}+\mathrm{C} _{2}}=\frac{\mathrm{KC} _{0}}{\mathrm{~K}+1} \quad \text { (in series) } \\ & \Rightarrow \mathrm{Q} _{2}=\mathrm{Q} _{1}=\mathrm{C} _{\mathrm{net}} \mathrm{V}=\left(\frac{\mathrm{KC} _{0}}{\mathrm{~K}+1}\right) \mathrm{V} \qquad\qquad \ldots . \ldots {(1)} \end{align*} $$

[Charge in series remains same]

When slab from $\mathrm{C} _{2}$ is removed, now

$$ \begin{aligned} & \mathrm{C} _{1}=\mathrm{C} _{0} \text { and } \mathrm{C} _{2}=\mathrm{C} _{0} \\ & C _{\text {net }}=\frac{C}{2} \\ & \text { (in series) } \end{aligned} $$

As $\mathrm{Q}$ is same on each

$$ \begin{equation*} \mathrm{Q} _{2}^{\prime}=\mathrm{Q} _{1}^{\prime}=\frac{\mathrm{C} _{0}}{2} \mathrm{~V} \qquad\qquad \ldots . \ldots {(2)} \end{equation*} $$

From (1) & (2)

$$ \frac{\mathrm{Q} _{2}^{\prime}}{\mathrm{Q} _{2}}=\frac{\mathrm{K}+1}{2 \mathrm{~K}} $$

Correct answer: (3)

Easy $\qquad \quad $ Redistribution of charges

25. A capacitor of $20 \mu \mathrm{F}$ and charged to $500 \mathrm{~V}$ is connected in parallel with another capacitor of $10 \mu \mathrm{F}$ charged to $200 \mathrm{~V}$. The common potential of the system is:

(1) $100 \mathrm{~V}$

(2) $120 \mathrm{~V}$

(3) $350 \mathrm{~V}$

(4) $400 \mathrm{~V}$

Show Answer

Correct Answer: (4)

Solution:

$\mathrm{V} _{\text {common }}=\frac{\mathrm{q} _{\text {net }}}{\mathrm{C} _{\text {net }}}$

or $\mathrm{V} _{\text {common }}=\frac{\mathrm{C} _{\mathrm{A}} \mathrm{V} _{\mathrm{A}}+\mathrm{C} _{\mathrm{B}} \mathrm{V} _{\mathrm{B}}}{\mathrm{C} _{\mathrm{A}}+\mathrm{C} _{\mathrm{B}}}$

or $\quad \mathrm{V} _{\text {common }}=\frac{20 \times 10^{-6} \times 500+10 \times 10^{-6} 200}{(20+10) \times 10^{-6}}$

or $\quad \mathrm{V} _{\text {common }}=\frac{12000}{30}=400 \mathrm{~V}$

Correct answer: (4)

Easy $\qquad \quad $ Redistribution of charges

26. Two conducting sphere $P$ and $Q$ of radius $r _{1}$ and $r _{2}$ are shown in the figure $\left(r _{1}>r _{2}\right)$

Both are positively charged and isolated from surroundings. They had been connected by a connecting wire but the wire is removed after some time.

$~$

Which of the following statements are true;

(a) Sphere $\mathrm{P}$ is at higher potential

(b) Both spheres are at same potential

(c) Magnitude of charge on $\mathrm{P}$ is greater than $\mathrm{Q}$

(d) Both the shperes will be having same charge after removal of wire

(1) (a) and (d)

(2) (b) and (d)

(3) (b) and (c)

(4) (c) and (a)

Show Answer

Correct Answer: (3)

Solution:

When both conducting spheres are connected by a wire, there will be redistribution of charge till they accquire the same potential (statement (b) is correct).

As $\mathrm{V} _{\mathrm{P}}=\mathrm{V} _{\mathrm{Q}}$

$$ \begin{aligned} & \frac{1}{4 \pi \varepsilon _{0}} \frac{\mathrm{Q} _{\mathrm{P}}}{\mathrm{r} _{1}}=\frac{1}{4 \pi \varepsilon _{0}} \frac{\mathrm{Q} _{\mathrm{Q}}}{\mathrm{r} _{2}} \\ & \Rightarrow \frac{\mathrm{Q} _{\mathrm{P}}}{\mathrm{Q} _{\mathrm{Q}}}=\frac{\mathrm{r} _{1}}{\mathrm{r} _{2}} \Rightarrow \mathrm{Q} _{\mathrm{P}}>\mathrm{Q} _{\mathrm{Q}}\left(\text { as } \mathrm{r} _{1}>\mathrm{r} _{2}\right) \quad \text { (statement c) } \end{aligned} $$

Correct answer: (3)

Difficult

Redistribution of Charges

27. Two capacitor $C _{1}=1 \mu \mathrm{F}$ and $\mathrm{C} _{1}=4 \mu \mathrm{F}$ are charged to a p.d. of $100 \mathrm{~V}$ and $200 \mathrm{~V}$ respectively. The charged capacitors are now connected to each other with the terminals of opposite sign connected together. Now the energy of the system is

(1) $0.012 \mathrm{~J}$

(2) $0.049 \mathrm{~J}$

(3) $0.081 \mathrm{~J}$

(4) $0.162 \mathrm{~J}$

Show Answer

Correct Answer: (2)

Solution:

When terminals of opposite sign one connected together, the magnitude of net charges finally is equal to the difference of charges before the connection.

Initial charge on $\mathrm{C} _{1}=\mathrm{Q} _{1}=\mathrm{C} _{1} \mathrm{~V} _{1}=100 \mu \mathrm{C}$

Initial charge on $\mathrm{C} _{2}=\mathrm{Q} _{2}=\mathrm{C} _{2} \mathrm{~V} _{2}=800 \mu \mathrm{C}$

$\mathrm{Q} _{\text {net }}=\mathrm{Q} _{2}-\mathrm{Q} _{1}=700 \mu \mathrm{C}$

$\mathrm{V} _{\text {common }}=\frac{\mathrm{Q} _{\text {net }}}{\mathrm{C} _{\text {net }}}=\frac{700 \times 10^{-6}}{5 \times 10^{-6}}=140 \mathrm{~V}$

The energy of system

$\mathrm{U}=\frac{1}{2} \mathrm{C} _{\text {net }} \mathrm{V} _{\text {common }}^{2}$

or $\quad \mathrm{U}=\frac{1}{2} \times 5 \times 10^{-6} \times(140)^{2}$

or $\mathrm{U}=0.049 \mathrm{~J}$

Correct answer: (2)

Average

Capacitor (Parallel-Plate)

28. The force exerted by each plate on the other plate of a parallel plate capacitor of capacitance ’ $C$ ’ and separation between the plates ’ $d$ ’ with potential difference ’ $V$ ’ between the plates is given by

(1) $\frac{\mathrm{CV}^{2}}{2 \mathrm{~d}}$

(2) $\frac{\mathrm{C}^{2} V^{2}}{2 d^{2}}$

(3) $\frac{C^{2} V^{2}}{d^{2}}$

(4) $\frac{\mathrm{dV}^{2}}{\mathrm{C}}$

Show Answer

Correct Answer: (2)

Solution:

Force exerted by each plate on the other plate of a parallel plate capacitor is given by

$$ \mathrm{F}=\frac{1}{2} \mathrm{qE} $$

where $\mathrm{q}$ is the magnitude of charge on either plate and $\mathrm{E}$, electric field between the plates.

Also $\mathrm{E}=\frac{\mathrm{V}}{\mathrm{d}}$ where $\mathrm{V}$ is the p.d. between the plates and ’ $\mathrm{d}$ ’ plate separation.

$$ \begin{aligned} & \Rightarrow F=\frac{1}{2} q \frac{V}{d} \\ & \text { or } F=\frac{1}{2} \frac{C^{2}}{d} \quad(\text { Putting } q=C V) \end{aligned} $$

Correct answer: (2)

Difficult

Combination of Capacitors

29. Five similar capacitor plates, each of area A, are placed at a equal distance ’ $d$ ’ apart and are connected to a source of e.m. E as shown in the following diagram. The charge on the plate 1 and 4 will be

$~$

(1) $\frac{\varepsilon _{0} \mathrm{~A}}{\mathrm{~d}}, \frac{-2 \varepsilon _{0} \mathrm{~A}}{\mathrm{~d}}$

(2) $\frac{\varepsilon _{0} \mathrm{AV}}{\mathrm{d}}, \frac{-2 \varepsilon _{0} \mathrm{AV}}{\mathrm{d}}$

(3) $\frac{-\varepsilon _{0} \mathrm{AV}}{\mathrm{d}}, \frac{-3 \varepsilon _{0} \mathrm{AV}}{\mathrm{d}}$

(4) $\frac{\varepsilon _{0} \mathrm{AV}}{\mathrm{d}}, \frac{-4 \varepsilon _{0} \mathrm{AV}}{\mathrm{d}}$

Show Answer

Correct Answer: (2)

Solution:

From equivalent circuit diagram, charge on first plate is

$$ \mathrm{Q}=\mathrm{CV} \Rightarrow \mathrm{Q}=\frac{\varepsilon _{0} \mathrm{AV}}{\mathrm{d}} $$

$~$

Charge on fourth plate $Q^{\prime}=(-C V) \Rightarrow Q^{\prime}=\frac{-\varepsilon _{0} A V}{d}$

As plate 4 is repeated twice, hence charge on 4 will be $Q^{\prime \prime}=2 Q^{\prime}$

i.e. $Q^{\prime \prime}=\frac{-2 \varepsilon _{0} A V}{d}$

Hence correct answer will be (2)

Average

Capacitor

30. A network of four capacitor of capacitance equal to $C _{1}=C, C _{2}=2 C, C _{3}=3 \mathrm{C}$ and $C _{4}=4 \mathrm{C}$ are connected to an ideal battery of $\mathrm{V}$ volts as shown in figure. The ratio of charges on $\mathrm{C} _{2}$ and $\mathrm{C} _{4}$ is

$~$

(1) $\frac{4}{7}$

(2) $\frac{3}{22}$

(3) $\frac{7}{4}$

(4) $\frac{22}{3}$

Show Answer

Correct Answer: (2)

Solution:

As in the given network $\mathrm{C} _{1}, \mathrm{C} _{2}$ and $\mathrm{C} _{3}$ are in series hence, equivalent of this series combination is

$\frac{1}{\mathrm{C}^{\prime}}=\frac{1}{\mathrm{C} _{1}}+\frac{1}{\mathrm{C} _{2}}+\frac{1}{\mathrm{C} _{3}}$

$\frac{1}{\mathrm{C}^{\prime}}=\frac{1}{\mathrm{C}}+\frac{1}{2 \mathrm{C}}+\frac{1}{3 \mathrm{C}}$

$\Rightarrow \mathrm{C}^{\prime}=\frac{6 \mathrm{C}}{11}$

This combination $\mathrm{C}^{\prime}$ is in parallel with $\mathrm{C} _{4}$.

$\Rightarrow \mathrm{C} _{\text {net }}=\mathrm{C}^{\prime}+\mathrm{C} _{4}=\frac{6 \mathrm{C}}{11}+4 \mathrm{C}$

$ =\frac{50}{11} \mathrm{C} $

$~$

In parallel combination ’ $V$ ’ remains same.'

$ \Rightarrow \frac{\mathrm{Q}^{\prime}}{\mathrm{C}^{\prime}}=\frac{\mathrm{Q} _{4}}{\mathrm{C} _{4}} $

or $\frac{\mathrm{Q}^{\prime}}{\mathrm{Q} _{4}}=\frac{\frac{6 \mathrm{C}}{11}}{4 \mathrm{C}}=\frac{3}{22}$

As Q’ is the charge on $\mathrm{C}^{\prime}$ (Combination of $\mathrm{C} _{1}, \mathrm{C} _{2}$ and $\mathrm{C} _{3}$ in series or change remains same in series. This means change on $\mathrm{C} _{2}$ will be same as change on $\mathrm{C}^{\prime}=\mathrm{Q}^{\prime}$ )

Hence, $\frac{\text { Charge on } \mathrm{C} _{2}}{\text { Charge on } \mathrm{C} _{4}}=\frac{3}{22}$

Correct answer: (2)

Average

Electric Flux

31. Consider an electric field $\mathbf{E}=\mathrm{E} _{0} \hat{\mathrm{i}}$, where $\mathrm{E} _{0}$ is a constant. The flux through the shaded area as shown in the figure, due to this field $\mathbf{E}$ is:

$~$

(1) $\mathrm{E} _{0} \mathrm{a}^{2}$

(2) $2 \mathrm{E} _{0} \mathrm{a}^{2}$

(3) $2 \sqrt{2} \mathrm{E} _{0} \mathrm{a}^{2}$

(4) $\sqrt{2} \mathrm{E} _{0} \mathrm{a}^{2}$

Show Answer

Correct Answer: (2)

Solution:

Flux, $\phi=\mathbf{E} \mathbf{A}$, for uniform field.

$$ \begin{aligned} & \mathrm{E} _{0} \times(\text { Area perpendicular to } \hat{\mathrm{i}}) \\ & =\mathrm{E} _{0} \times(2 \mathrm{a} \times \mathrm{a}) \\ & =2 \mathrm{E} _{0} \mathrm{a}^{2} \end{aligned} $$

Correct answer is $2 \mathrm{E} _{0} \mathrm{a}^{2}$

Difficult

Electric Fied

32. Two oppositely charged non-conducting solid spheres, of radii $R$ and $\frac{3}{2} R$, having uniform volume change densitites $\rho _{1}$, and $\rho _{2}$ respectively, touch each other. The net electric field at a distance $2 R$ from the centre of the smaller sphere, along the line joining the centres of the spheres, is zero. The ratio $\frac{\rho _{1}}{\rho _{2}}$ is equal to:

(1) -4

(2) $-\frac{2}{3}$

(3) $\frac{2}{3}$

(4) 2

Show Answer

Correct Answer: (2)

Solution:

$~$

Since the spheres are oppositely charged, the point $\mathrm{P}$, where the net field is zero, should lie on the left side of smaller sphere, as shown in the figure.

Let $E _{1}$ be the field due to smaller sphere at $P$ and $E _{2}$ be that due to the bigger one, at $P . E _{1}$ and $E _{2}$ should be equal and opposite.

Now, $\mathrm{E} _{1}=\frac{1}{4 \pi \varepsilon _{0}} \frac{\frac{4}{3} \pi \mathrm{R}^{3} \rho _{1}}{(2 \mathrm{R})^{2}}$ and $\mathrm{E} _{2}=\frac{1}{4 \pi \varepsilon _{0}} \frac{\frac{4}{3} \pi\left(\frac{3}{2} \mathrm{R}\right)^{3} \rho _{2}}{\left(\frac{9}{2} \mathrm{R}\right)^{2}}$

$\therefore \frac{1}{4 \pi \varepsilon _{0}} \frac{\frac{4}{3} \pi \mathrm{R}^{3} \rho _{1}}{4 \mathrm{R}^{2}}=\frac{1}{4 \pi \varepsilon _{0}} \frac{\frac{4}{3} \pi \frac{3^{3}}{2^{3}} \mathrm{R}^{3} \rho _{2}}{\frac{9^{2}}{2^{2}} \mathrm{R}^{2}}$

or $\quad \frac{\rho _{1}}{4}=\frac{3^{3}}{2^{3}} \times \frac{2^{2}}{3^{4}} \rho _{2}$

or $\quad \frac{\rho _{2}}{\rho _{1}}=\frac{4}{6}=\frac{2}{3}$

As one of the charge is negative,

$ \frac{\rho _{1}}{\rho _{2}}=-\frac{-2}{3} $

Correct answer is $\frac{\rho _{1}}{\rho _{2}}=-\frac{-2}{3}$

Difficult

Electric Potential Energy

33. Three piont charges $q, 2 q$ and $8 q$ Coulomb’s are to be placed on $x$-axis, such that the maximum distance between any two is $9 \mathrm{~cm}$ and the potential energy of the system is the minimum. If $q$ is to be placed at the origin, the locations of $2 q$ and $8 q$ are respectively at:

(1) $6 \mathrm{~cm}$ and $9 \mathrm{~cm}$

(2) $-3 \mathrm{~cm}$ and $6 \mathrm{~cm}$

(3) $6 \mathrm{~cm}$ and $-3 \mathrm{~cm}$

(4) $-4 \mathrm{~cm}$ and $5 \mathrm{~cm}$

Show Answer

Correct Answer: (2)

Solution:

For minimum potential energy, the charges of maximum energy should be at farthest position. Therefore the charges of $2 \mathrm{q}$ and $8 \mathrm{q}$ should be $9 \mathrm{~cm}$ apart (as $9 \mathrm{~cm}$ is the maximum possible according to the question). Let the positions of the three charges be as shown in the figure.

$~$

Here, $\mathrm{q}$ is at the origin.

$2 \mathrm{q}$ at $x$ to the left of origin.

$8 \mathrm{q}$ at $(9-x)$ to the right of the origin.

The potential energy of this system is:

$\mathrm{U}=\frac{1}{4 \pi \varepsilon _{0}}\left[\frac{2 \mathrm{q} \times \mathrm{q}}{x}+\frac{2 \mathrm{q} \times 8 \mathrm{q}}{9}+\frac{\mathrm{q} \times 8 \mathrm{q}}{(9-x)}\right]$

or $\quad \mathrm{U}=\frac{2 \mathrm{q}^{2}}{4 \pi \varepsilon _{0}}\left[\frac{1}{x}+\frac{8}{9}+\frac{4}{(9-x)}\right]$

But, for $\mathrm{U}$ to be minimum,

$ \frac{\mathrm{dU}}{\mathrm{d} x}=0 $

$\therefore \frac{2 \mathrm{q}^{2}}{4 \pi \varepsilon _{0}}\left[-\frac{1}{x^{2}}+0+\frac{4}{(9-x)^{2}}\right]=0$, for $\mathrm{U}=\mathrm{U} _{\text {min. }}$

Hence,

$$ \frac{1}{x^{2}}=\frac{4}{(9-x)^{2}} $$

or $4 x^{2}=(9-x)^{2}$

or $\quad \pm 2 x=9-x$

or $\quad x=3 \mathrm{~cm}$, to the left of origin

or $\quad x=-9 \mathrm{~cm}$, or $9 \mathrm{~cm}$ to the right of origin, which is not possible, here.

$\therefore 2 q$ is to be placed at $-3 \mathrm{~cm}$ with respect to the origin and $8 \mathrm{q}$ should be placed at $+6 \mathrm{~cm}$ from the origin, on the $x$-axis.

The correct answer is:

$2 \mathrm{q}$ at $-3 \mathrm{~cm}$

$q$ at origin

$8 \mathrm{q}$ at $+6 \mathrm{~cm}$

Correct answer: (2)

Average

Electric Potential

34. When the intensity of electric field becomes $3 \times 10^{6} \mathrm{Vm}^{-1}$, then the insulation of air is broken down. A metallic spere in air is able to store a maximum charge of $30 \mu \mathrm{C}$. The maximum potential the sphere can have is:

(1) $90 \mathrm{~V}$

(2) $2.7 \times 10^{6} \mathrm{~V}$

(3) $1.8 \times 10^{6} \mathrm{~V}$

(4) $0.9 \times 10^{6} \mathrm{~V}$

Show Answer

Correct Answer: (4)

Solution:

For the maximum values,

$ \begin{aligned} & \mathrm{E}=3 \times 10^{6} \mathrm{Vm}^{-1} \\ & \mathrm{Q}=30 \mu \mathrm{C}=30 \times 10^{-6} \mathrm{C} \end{aligned} $

But, $\mathrm{E}=\frac{1}{4 \pi \varepsilon _{0}} \frac{\mathrm{Q}}{\mathrm{R}^{2}} \quad \quad\quad \quad (\mathrm{R}=\text { radius of the sphere })$

$\therefore \mathrm{R}^{2}=\frac{1}{4 \pi \varepsilon _{0}} \times \frac{\mathrm{Q}}{\mathrm{E}}$

$ =9 \times 10^{9} \times \frac{30 \times 10^{-6}}{3 \times 10^{6}} $

$\mathrm{R}^{2}=9 \times 10^{-2} \mathrm{~m} _{2}$

$\therefore \mathrm{R}=3 \times 10^{-1} \mathrm{~m}$

$=0.3 \mathrm{~m}$

Therefore, the maximum potential it can have is

$ \begin{gathered} \mathrm{V}=\frac{1}{4 \pi \varepsilon _{0}} \cdot \frac{\mathrm{Q}}{\mathrm{R}}=\mathrm{E} \times \mathrm{R} \\ =3 \times 10^{6} \times 0.3 \\ =0.9 \times 10^{6} \mathrm{~V} \end{gathered} $

Correct answer: (4)

Difficult

Electric Potential

35. A spherical metal shell $A$ of radius $R _{A}$ and a solid metal sphere $B$ of radius $R _{B}\left(<R _{A}\right)$ are kept at a large distance and are given charges $Q$ each. They are then connected by a thin metal wire. Which of the following statements are correct regarding the electrical conditions of A and B, after they are connected with the wire?

(i) $\mathrm{E} _{\mathrm{A}}^{\text {inside }}=\mathrm{E} _{\mathrm{B}}^{\text {inside }}=0$

(ii) $\frac{\sigma _{\mathrm{A}}}{\sigma _{\mathrm{B}}}=\frac{\mathrm{R} _{\mathrm{A}}}{\mathrm{R} _{\mathrm{B}}}$

(iii) $\mathrm{Q} _{\mathrm{A}}>\mathrm{Q} _{\mathrm{B}}$

(iv) $\mathrm{E} _{\mathrm{A}}^{\text {onsurface }}=\mathrm{E} _{\mathrm{B}}^{\text {on surface }}$

(1) (i), (ii) and (iii)

(2) (i), (iii) and (iv)

(3) (ii) and (iv)

(4) (i), (ii), (iii) and (iv)

Show Answer

Correct Answer: (2)

Solution:

For metal sphere or shell, charges reside only on the surface. As there is no charge inside, by using Gauss’s theorem, we can show that $\mathrm{E} _{\mathrm{A}}^{\text {inside }}=\mathrm{E} _{\mathrm{B}}^{\text {inside }}=0$.

So statement (i) is correct.

Let the charges on $A$ and $B$, after they are connected with the wire, be $Q _{A}$ and $Q _{B}$ respectively.

$~$

The final potential of $A$ and $B$ will be equal, as charges flow from $B$ and $A$, till the potential become equal.

Therefore $\mathrm{V} _{\mathrm{A}}=\mathrm{V} _{\mathrm{B}}$

So, $\frac{1}{4 \pi \varepsilon _{0}} \frac{\mathrm{Q} _{\mathrm{A}}}{\mathrm{R} _{\mathrm{A}}}=\frac{1}{4 \pi \varepsilon _{0}} \frac{\mathrm{Q} _{\mathrm{B}}}{\mathrm{R} _{\mathrm{B}}}$

or $\frac{Q _{A}}{R _{B}}=\frac{Q _{A}}{R _{B}}$

Since $\mathrm{R} _{\mathrm{A}}>\mathrm{R} _{\mathrm{B}}, \mathrm{Q} _{\mathrm{A}}>\mathrm{Q} _{\mathrm{B}}$

So, optino (3) gives a correct statement.

Now, $\mathrm{Q} _{\mathrm{A}}=4 \pi \mathrm{R} _{\mathrm{A}}^{2} \times \sigma _{\mathrm{A}}$

$ \mathrm{Q} _{\mathrm{B}}=4 \pi \mathrm{R} _{\mathrm{A}}^{2} \times \sigma _{\mathrm{B}} $

$\therefore \frac{\mathrm{Q} _{\mathrm{A}}}{\mathrm{Q} _{\mathrm{B}}}=\frac{\mathrm{R} _{\mathrm{A}}^{2}}{\mathrm{R} _{\mathrm{B}}^{2}} \frac{\sigma _{\mathrm{A}}}{\sigma _{\mathrm{B}}}$

or $\quad \frac{\mathrm{R} _{\mathrm{A}}}{\mathrm{R} _{\mathrm{B}}}=\frac{\mathrm{R} _{\mathrm{A}}^{2}}{\mathrm{R} _{\mathrm{B}}^{2}} \frac{\sigma _{\mathrm{A}}}{\sigma _{\mathrm{B}}}$

$\therefore \frac{\sigma _{\mathrm{A}}}{\sigma _{\mathrm{B}}}=\frac{\mathrm{R} _{\mathrm{B}}}{\mathrm{R} _{\mathrm{A}}}$

So, option (1) is an incorrect statement.

Electric field $\mathrm{E}$, just outside the surface of $\mathrm{A}$ and $\mathrm{B}$ are:

$\mathrm{E} _{\mathrm{A}}=\frac{\sigma _{\mathrm{A}}}{\varepsilon _{0}}$ and $\mathrm{E} _{\mathrm{B}}=\frac{\sigma _{\mathrm{B}}}{\varepsilon _{0}}$

$\therefore \frac{\mathrm{E} _{\mathrm{A}}}{\mathrm{E} _{\mathrm{B}}}=\frac{\sigma _{\mathrm{A}}}{\sigma _{\mathrm{B}}}$

But, $\frac{\sigma _{\mathrm{A}}}{\sigma _{\mathrm{B}}}=\frac{\mathrm{R} _{\mathrm{B}}}{\mathrm{R} _{\mathrm{A}}}$

$\therefore \frac{\mathrm{E} _{\mathrm{A}}}{\mathrm{E} _{\mathrm{B}}}=\frac{\mathrm{R} _{\mathrm{B}}}{\mathrm{R} _{\mathrm{A}}}$ or $\mathrm{E} _{\mathrm{A}}<\mathrm{E} _{\mathrm{B}} \quad\left(\right.$ as $\left.\mathrm{R} _{\mathrm{B}}<\mathrm{R} _{\mathrm{A}}\right)$

Hence, option (4) is a correct statement.

Difficult

Electric Force

36. A tiny spherical oil drop carrying a net charge $q$ is balanced is still air with a vertically upward uniform electric field of strength $\frac{81 \pi}{7} \times 10^{5} \mathrm{~V}-\mathrm{m}^{-1}$. When the field is switched off, this oil drop is observed to fall with a terminal velocity $2 \times 10^{-3} \mathrm{~ms}^{-1}$. If the density of air is neglected and taking $\mathrm{g}=9.8 \mathrm{~ms}^{-2}$, viscosity of air $=1.8 \times 10^{-5} \mathrm{~N}-\mathrm{S}-\mathrm{m}^{-2}$, density of oil $=900$ $\mathbf{k g m}^{-3}$, the number of electrons missing from the oil drop is:

(1) 2

(2) 3

(3) 4

(4) 5

Show Answer

Correct Answer: (4)

Solution:

Since the drop is balanced is a vertically upward electric field, the change q on it should be positive and so few electrons will be missing from it.

$$ \begin{aligned} & \mathrm{q}=+\mathrm{ne} \\ & \mathrm{n}=? \end{aligned} $$

We find qusing the following equations.

When the drop is balanced,

$$ \mathrm{qE}=\mathrm{mg} \quad(\text { Density of air is negected }) $$

or $\quad \mathrm{qE}=\frac{4}{3} \pi \mathrm{r}^{3} \rho \mathrm{g} \qquad \quad \quad \ldots . \ldots (1) $

When it falls with terminal velocity $\mathrm{v} _{\mathrm{t}}, \mathrm{a}=0$ and so net force on it is again zero.

Since field is switched off, we have,

$6 \pi \eta \mathrm{rv} _{\mathrm{t}}=\frac{4}{3} \pi \mathrm{r}^{3} \rho \mathrm{g} \qquad \quad \quad \ldots . \ldots (2)$

$\therefore \mathrm{r}^{2}=\frac{9}{2} \frac{\eta \mathrm{v} _{\mathrm{t}}}{\rho \mathrm{g}}$

or $\quad r=\left(\frac{9}{2} \frac{\eta v _{t}}{\rho g}\right)^{1 / 2} \qquad \quad \quad \ldots . \ldots (3) $

From (1) and (2), we have

$\mathrm{qE}=6 \pi \eta \mathrm{rv} _{\mathrm{t}}$

or $\quad q=\frac{6 \pi \eta v _{t}}{E} \times r$

$ \begin{aligned} & =\frac{6 \pi \eta v _{t}}{\mathrm{E}} \times\left(\frac{9}{2} \frac{\eta v _{t}}{\rho g}\right)^{1 / 2} \\ & =\frac{6 \pi \times 1.8 \times 10^{-5} \times 2 \times 10^{-3}}{\frac{81 \pi}{7} \times 10^{5}} \times\left(\frac{9}{2} \times \frac{1.8 \times 10^{-5} \times 2 \times 10^{-3}}{900 \times 9.8}\right)^{1 / 2} \\ & =\frac{0.8 \times 7}{3} \times 10^{-13} \times\left(\frac{1.8}{9.8} \times 10^{-10}\right)^{1 / 2} \\ & =\frac{0.8 \times 7}{3} \times 10^{-13} \times\left(\frac{9}{49} \times 10^{-10}\right)^{1 / 2} \\ & =\frac{0.8 \times 7}{3} \times 10^{-13} \times \frac{3}{7} \times 10^{-5} \\ & =0.8 \times 10^{-18} \\ & =8 \times 10^{-19} \mathrm{C} \end{aligned} $

But $q=$ ne

$\therefore \mathrm{n}=\frac{\mathrm{q}}{\mathrm{e}}=\frac{8 \times 10^{-19} \mathrm{C}}{1.6 \times 10^{-19} \mathrm{C}}$

$=5$

Correct answer: (4)

Difficult

Capacitor with Dielectric

37. The capacitance of parallel plate capacitor is $50 \mathrm{pF}$ and the distance between the plates is 4 $\mathrm{mm}$. It is charged to $200 \mathrm{~V}$, after that the chargnig battery is removed. Now, a dielectric slab of thickness $2 \mathrm{~mm}$ and dielectric constant 4 is placed between the plates. The exess heat produced in the slab, during this process, is;

(1) $1.0 \times 10^{-6} \mathrm{~J}$

(2) $6.25 \times 10^{-7} \mathrm{~J}$

(3) $3.75 \times 10^{-7} \mathrm{~J}$

(4) $2.50 \times 10^{-7} \mathrm{~J}$

Show Answer

Correct Answer: (3)

Solution:

With air as dielectric, the capacitance of a parallel plate capacitor is;

$$ \mathrm{C} _{0}=\frac{\varepsilon _{0} \mathrm{~A}}{\mathrm{~d}}=\frac{\varepsilon _{0} \mathrm{~A}}{4} $$

Now, with a dielectric slab of thickness $t$ and dielectric constant $K$, the capacitance is:

$$ C=\frac{\varepsilon _{0} A}{d-t+(t / k)} $$

When $\mathrm{C} _{0}$ is connected to the battery, then

$$ \begin{aligned} \mathrm{Q}=\mathrm{C} _{0} \mathrm{~V} _{0}= & 50 \times 10^{-12} \times 200 \\ = & 10^{-8} \mathrm{C} \end{aligned} $$

After charging the battery is removed. Hence the change Q will remain unchanged, there after. How, suppose the capacitance changes to $\mathrm{C}$, after placing the dielectric.

Here, $\mathrm{C}=\frac{\varepsilon _{0} \mathrm{~A}}{\mathrm{~d}-\mathrm{t}+(\mathrm{t} / \mathrm{K})}$

$$ \begin{aligned} & =\frac{\varepsilon _{0} \mathrm{~A}}{4-2+\frac{2}{4}} \quad(\text { As } \mathrm{t}=2 \mathrm{~m}, \mathrm{~d}=3 \mathrm{~mm} \text { and } \mathrm{K}=4) \\ & =\frac{\varepsilon _{0} \mathrm{~A}}{10} \times 4 \end{aligned} $$

Since $\mathrm{Q}$ remains same, $\mathrm{Q}=\mathrm{C} _{0} \mathrm{~V} _{0}=\mathrm{CV}$ ( $\mathrm{V}$ is the new potential) $\therefore \mathrm{V}=\frac{\mathrm{C} _{0} \mathrm{~V} _{0}}{\mathrm{C}}$

$ \begin{aligned} & =\frac{\varepsilon _{0} \mathrm{~A} \times 200}{4 \times \frac{\varepsilon _{0} \mathrm{~A} \times 4}{10}} \\ & =125 \mathrm{~V} \end{aligned} $

Now, energy loss, when the slab is placed

$ \begin{aligned} \Delta \mathrm{U} & =\mathrm{U} _{\mathrm{i}}-\mathrm{U} _{\mathrm{f}} \\ & =\frac{1}{2} \mathrm{QV} _{0}-\frac{1}{2} \mathrm{QV} \\ & =\frac{1}{2} \mathrm{Q}\left(\mathrm{V} _{0}-\mathrm{V}\right) \\ & =\frac{1}{2} \mathrm{Q}(200-125) \\ & =\frac{1}{2} \times 10^{-8} \times 75 \\ & =37.5 \times 10^{-8} \\ & =3.75 \times 10^{-7} \mathrm{~s} \end{aligned} $

This energy loss will appear as surplus heat produced in the slab. So heat produced in the slab $=3.75 \times 10^{-7} \mathrm{~J}$ Correct answer is (3)

Difficult

Combination of Capacitor

38. In the given network, $C _{1}=1 \mu \mathrm{F}, \mathrm{C} _{2}=2 \mu \mathrm{F}, \mathrm{C} _{3}=3 \mu \mathrm{F}$ and $\mathrm{C} _{4}=4 \mu \mathrm{F}$. When the key is closed $-q$ charge flows from $A$ to $B$. The numerical value of $q$ in $\mu \mathrm{C}$ is:

$~$

(1) 2.4

(2) 1.8

(3) 0.8

(4) 0.2

Show Answer

Correct Answer: (1)

Solution:

When $\mathrm{K}$ is open, the network will be as shown in the figure. $\mathrm{C} _{1}$ and $\mathrm{C} _{3}$ are in series. Similarly $\mathrm{C} _{2}$ and $\mathrm{C} _{4}$ are in series. Both the series combination get $12 \mathrm{~V}$ each.

$\mathrm{Q} _{1}=\mathrm{Q} _{3}=\mathrm{V} \frac{\mathrm{C} _{1} \mathrm{C} _{3}}{\mathrm{C} _{1}+\mathrm{C} _{3}}=12 \times \frac{1 \times 3}{1+3}=9 \mu \mathrm{C}$

$\mathrm{Q} _{2}=\mathrm{Q} _{4}=\mathrm{V} \frac{\mathrm{C} _{2} \mathrm{C} _{4}}{\mathrm{C} _{2}+\mathrm{C} _{4}}=12 \times\left(\frac{2 \times 4}{2+4}\right)=16 \mu \mathrm{C}$

$\therefore \mathrm{Q}=9 \mu \mathrm{C}+16 \mu \mathrm{C}=25 \mu \mathrm{C}$

$~$

Now, When $\mathrm{K}$ is closed, the network will be as shwon in the figure.

Here, $\mathrm{C} _{1}$ and $\mathrm{C} _{2}$ are in parallel. $\mathrm{C} _{3}$ and $\mathrm{C} _{4}$ are in parallel and these parallel combinations are in series.

Here, the effective capactitance $\mathrm{C}$ is:

$$ \begin{aligned} C= & \frac{\left(C _{1}+C _{2}\right) \times\left(C _{3}+C _{4}\right)}{C _{1}+C _{2}+C _{3}+C _{4}}=\frac{(1+2) \times(3+4)}{1+2+3+4} \\ & =\frac{3 \times 7}{10}=2.1 \mu \mathrm{F} \end{aligned} $$

$~$

$\therefore$ Change stored in $\mathrm{Q}=\mathrm{CV}=2.1 \times 12=25.2 \mu \mathrm{C}$.

So, chnage on $\left(\mathrm{C} _{1}+\mathrm{C} _{2}\right)=3 \mu \mathrm{F}$ is $\mathrm{q} _{1}+\mathrm{q} _{2}=25.2 \mu \mathrm{C}$.

Hence, potential difference across $\mathrm{C} _{1}$, which is equal to that across $\mathrm{C} _{2}$ is:

$\mathrm{V} _{1}=\mathrm{V} _{2}=\frac{25.2 \mu \mathrm{C}}{3 \mu \mathrm{F}}=8.4 \mathrm{~V}$

Similary, $\mathrm{V} _{3}=\mathrm{V} _{4}=3.6 \mathrm{~V}$

Now, $\mathrm{q} _{1}=\mathrm{C} _{1} \mathrm{~V} _{1}=1 \times 8.4=8.4 \mu \mathrm{C}$

$ \begin{aligned} & \mathrm{q} _{2}=\mathrm{C} _{2} \mathrm{~V} _{2}=2 \times 8.4=16.8 \mu \mathrm{C} \\ & \mathrm{q} _{3}=\mathrm{C} _{3} \mathrm{~V} _{3}=3 \times 3.6=10.8 \mu \mathrm{C} \\ & \mathrm{q} _{4}=\mathrm{C} _{4} \mathrm{~V} _{4}=4 \times 3.6=14.4 \mu \mathrm{C} \end{aligned} $

Initial change on the $-\mathrm{ve}$ plate of $\mathrm{C} _{1}=-9 \mu \mathrm{C}$

Final change on the $-\mathrm{ve}$ plate of $\mathrm{C} _{1}=-8.4 \mu \mathrm{C}$

Change flowing down through $\mathrm{A}$, from $\mathrm{C} _{1}=-0.6 \mu \mathrm{C}$

Now, initial change on $+v e$ plate of $\mathrm{C} _{3}=+9 \mu \mathrm{C}$

Final change on + ve plate of $\mathrm{C} _{3}=+10.8 \mu \mathrm{C}$

$\therefore$ Chnage flowing down through $\mathrm{A}$, from $\mathrm{C} _{3}=-1.8 \mu \mathrm{C}$

Hence, total change flowing down through $\mathrm{A}=-0.6+-1.8$

$ =-2.4 \mu \mathrm{C} $

$\therefore$ Value of q in $\mu \mathrm{C}=2.4$

Correct answer: (1)

Difficult

Combination of Capacitor

39. A, B, C, and $D$ are four similar metallic parallel plates with separtions $d _{1}, d _{2}$, and $d _{3}$ respectively, such that $d _{1}: d _{2}: d _{3}=$ $1: 2: 3$. A source of potential $V$ is connected as shwon. The potential of A, B, C and D are respectively.

$~$

(1) $\mathrm{V}, \frac{2}{3} \mathrm{~V}, \frac{1}{3} \mathrm{~V}$ and 0

(2) $\mathrm{V}, \frac{2}{3} \mathrm{~V}, \frac{1}{6} \mathrm{~V}$ and 0

(3) $\mathrm{V}, \frac{6}{6} \mathrm{~V}, \frac{1}{2} \mathrm{~V}$ and 0

(4) $\mathrm{V}, \frac{1}{2} \mathrm{~V}, \frac{1}{6} \mathrm{~V}$ and 0

Show Answer

Correct Answer: (3)

Solution:

The arrangement is equivalent to three capacitors of $\mathrm{C} _{1}, \mathrm{C} _{2}$ and $\mathrm{C} _{3}$ connected in series. Plate area, $\mathrm{A}$, is same for all and for $\mathrm{C} _{1}, \mathrm{~d} _{1}=\mathrm{d}$; for $\mathrm{C} _{2}, \mathrm{~d} _{2}=2 \mathrm{~d}$ and for $\mathrm{C} _{3}, \mathrm{~d} _{3}=3 \mathrm{~d}$.

$~$

$ \begin{aligned} & \therefore \mathrm{C} _{1}=\frac{\varepsilon _{0} \mathrm{~A} _{1}}{\mathrm{~d} _{1}}=\frac{\varepsilon _{0} \mathrm{~A}}{\mathrm{~d}}=\mathrm{C} _{0} \\ & \mathrm{C} _{2}=\frac{\varepsilon _{0} \mathrm{~A} _{2}}{\mathrm{~d} _{2}}=\frac{\varepsilon _{0} \mathrm{~A}}{2 \mathrm{~d}}=\frac{\mathrm{C} _{0}}{2} \\ & \mathrm{C} _{3}=\frac{\varepsilon _{0} \mathrm{~A} _{3}}{\mathrm{~d} _{3}}=\frac{\varepsilon _{0} \mathrm{~A}}{3 \mathrm{~d}}=\frac{\mathrm{C} _{0}}{3} \end{aligned} $

If $C$ is the effective capacitance, then:

$ \frac{1}{\mathrm{C}}=\frac{1}{\mathrm{C} _{1}}+\frac{1}{\mathrm{C} _{2}}+\frac{1}{\mathrm{C} _{3}} $

or $\quad \frac{1}{\mathrm{C}}=\frac{1}{\mathrm{C} _{0}}+\frac{2}{\mathrm{C} _{0}}+\frac{3}{\mathrm{C} _{0}}$

or $\frac{1}{\mathrm{C}}=\frac{6}{\mathrm{C} _{0}}$

or $\mathrm{C}=\frac{\mathrm{C} _{0}}{6}$

The charge stored in system is:

$ \mathrm{q}=\mathrm{CV}, \mathrm{q}=\frac{\mathrm{C} _{0}}{6} \mathrm{~V} $

q is same for all.

$\therefore$ Potential difference for $\mathrm{C} _{1}$ is

$ \mathrm{V} _{1}=\frac{\mathrm{q}}{\mathrm{C} _{1}}=\frac{\mathrm{C} _{0} \mathrm{~V}}{\mathrm{C} _{0} \times 6}=\frac{\mathrm{V}}{6} $

Potential difference for $\mathrm{C} _{2}$ is

$ \mathrm{V} _{2}=\frac{\mathrm{q}}{\mathrm{C} _{2}}=\frac{\mathrm{C} _{0} \mathrm{~V}}{6 \times\left(\frac{\mathrm{C} _{0}}{2}\right)}=\frac{\mathrm{V}}{3} $

Potential difference for $\mathrm{C} _{3}$ is

$ \mathrm{V} _{3}=\frac{\mathrm{q}}{\mathrm{C} _{3}}=\frac{\mathrm{C} _{0} \mathrm{~V}}{6 \times\left(\frac{\mathrm{C} _{0}}{3}\right)}=\frac{\mathrm{V}}{2} $

Now, from the figure given: $\mathrm{V} _{\mathrm{A}}=\mathrm{V}$ and $\mathrm{V} _{\mathrm{D}}=0 \qquad \quad \qquad $ (as $\mathrm{D}$ is earthed)

Now, $\mathrm{V} _{\mathrm{A}}-\mathrm{V} _{\mathrm{B}}=\frac{\mathrm{V}}{6} \quad\left(\right.$ for $\left.\mathrm{C} _{1}\right)$

$\therefore \mathrm{V} _{\mathrm{B}}=\mathrm{V} _{\mathrm{A}}-\frac{\mathrm{V}}{6}=\mathrm{V}-\frac{\mathrm{V}}{6}=\frac{5}{6} \mathrm{~V}$

$\mathrm{V} _{\mathrm{B}}-\mathrm{V} _{\mathrm{C}}=\frac{\mathrm{V}}{3} \qquad \quad \quad $ $\left(\right.$ for $\left.\mathrm{C} _{2}\right)$

$\therefore \mathrm{V} _{\mathrm{C}}=\mathrm{V} _{\mathrm{B}}-\frac{\mathrm{V}}{3}$

$=5 \frac{\mathrm{V}}{6}-\frac{\mathrm{V}}{3}=\frac{3}{6} \mathrm{~V}=\frac{\mathrm{V}}{2}$

or $\quad \mathrm{V} _{\mathrm{A}}=\mathrm{V}, \mathrm{V} _{\mathrm{B}}=\frac{5}{6} \mathrm{~V}, \mathrm{~V} _{\mathrm{C}}=\frac{\mathrm{V}}{2}$ and $\mathrm{V} _{\mathrm{D}}=0$

Correct answer: (3)

Average

Electric field due to Spherical Body

40. Consider a sphere of radius $R$ with a volume distribution of charge represented by: $\rho _{(\mathrm{r})}=\mathrm{Kr}$ for $r \leq R$ and $\rho=0$ for $r \geq R$. The electric field intensity at any point where $r>R$ is given as:

(1) $\mathrm{E}=\frac{\mathrm{Kr}^{2}}{4 \varepsilon _{0}}$

(2) $\mathrm{E}=\frac{\mathrm{KR}^{4}}{4 \varepsilon _{0} \mathrm{r}^{2}}$

(3) $\mathrm{E}=\frac{\mathrm{KR}^{2}}{4 \varepsilon _{0}}$

(4) $\mathrm{E}=\frac{\mathrm{KR}^{4}}{\varepsilon _{0} \mathrm{r}^{2}}$

Show Answer

Correct Answer: (2)

Solution:

Let $S$ be a spherical Gaussian surface passing through the point, with $r>R$.

$\oint _{\mathrm{S}} \mathbf{E} \cdot \mathbf{d s}=\mathrm{E} \times 4 \pi \mathrm{r}^{2} \quad \qquad \ldots . \ldots (1) $

Now, the total charge of the sphere of radius $\mathrm{R}$ is;

$~$

$\mathrm{q}=\int _{\mathrm{r}=0}^{\mathrm{R}} \rho(\mathrm{r}) \mathrm{dV}$

or $\quad q=\int _{r=0}^{R} K r .4 \pi r^{2} d r$

or $\quad q=4 \pi K \frac{R^{4}}{4} \quad \qquad \ldots . \ldots (2)$

By Gauss’s theorem,

$ \begin{aligned} \oint _{S} \mathbf{E} \cdot \mathbf{d s}=\frac{1}{\varepsilon _{0}} \quad(\text { Charge enclosed by } S) \qquad \qquad \ldots . \ldots {(3)} \end{aligned} $

From(1), (2) and (3), we have:

$\mathrm{E} \times 4 \pi \mathrm{r}^{2}=\frac{1}{\varepsilon _{0}} \cdot 4 \pi \frac{\mathrm{KR}^{2}}{4}$

$\therefore \mathrm{E}=\frac{\mathrm{KR}^{4}}{4 \varepsilon _{0} \mathrm{r}^{2}}$

Correct answer: (2)

Average

Electric Field due to a Charged Shell

41. Two concentric conducting thin spherical shells $A$ and $B$ having radii $r _{A}$ and $r _{B}\left(r _{B}>r _{A}\right)$ are charged to $Q _{A}=+q$ and $Q _{B}=-2 q$ respectively. The variation the electric field (E) with distance $r$ from the common centre is repressented by the graph.

$~$

Show Answer

Correct Answer: (4)

Solution:

As the shells are made of conductors, charges reside only on their surfaces. Therefore electric field inside the shell, due to the charge on its surface will be zero.

Therefor, inside shell A, there no field due to charge on A or charge on B. In the space between A and B, the field is due to charge on A only. Since charge on in $Q _{A}=+q$, the field in positive direction of $r$.

Now, outside shell B, the charges on A and B behave like point charges at the common centre.

$\mathrm{Q} _{0}=+\mathrm{q}+-2 \mathrm{q}$

$=-\mathrm{q}$

For $r _{A}>r>r _{B}$, field is due to $+q$ only, as if it is at 0 .

$\therefore \mathrm{E}=\frac{\mathrm{Kq}}{\mathrm{r}^{2}}$

or $\quad \mathrm{E} \propto \frac{1}{\mathrm{r}^{2}}$

$~$

Therefore, $\mathrm{E}$ is positive and it varies inversely as square of $\mathrm{r}$, for $\mathrm{r} _{\mathrm{A}}<\mathrm{r}<\mathrm{r} _{\mathrm{B}}$.

For $r>r _{B}$, field is due to $-q$, as if it is at 0 .

$\therefore \mathrm{E}=-\frac{\mathrm{Kq}}{\mathrm{r}^{2}}$

Therefore, $\mathrm{E}$ is -ve and it also varies inversely as square of $\mathrm{r}$, for $\mathrm{r}>\mathrm{r} _{\mathrm{B}}$.

$\therefore \mathrm{E}=0$, up to $\mathrm{r}=\mathrm{r} _{\mathrm{A}}$

$E$ is positive and $E \alpha \frac{1}{r^{2}}$ for $r _{A}<r<r _{B}$.

$E$ is negative and $E \alpha \frac{1}{r^{2}}$ for $r>r _{B}$.

Correct graph is (4).

Easy

Potential

42. The potential of the electr ic field at any point $(x, y, z)$ is given by: $V=3 x^{2}+5$, where $x, y$ and $z$ are in metres and $V$ is in volts. The intensity of electric field at $(-2,1,0) \mathrm{m}$ is:

(1) $7 \mathrm{~V}-\mathrm{m}^{-1}$

(2) $-17 \mathrm{~V}-\mathrm{m}^{-1}$

(3) $12 \mathrm{~V}-\mathrm{m}^{-1}$

(4) $-12 \mathrm{~V}-\mathrm{m}^{-1}$

Show Answer

Correct Answer: (3)

Solution:

$\mathbf{E}=-\left(\frac{\partial \mathrm{V}}{\partial x} \hat{\mathrm{i}}+\frac{\partial \mathrm{V}}{\partial \mathrm{y}} \hat{\mathrm{j}}+\frac{\partial \mathrm{V}}{\partial \mathrm{z}} \hat{\mathrm{k}}\right)$

$ \begin{aligned} & \quad=-(6 x \hat{\mathrm{i}}+0 \hat{\mathrm{j}}+0 \hat{\mathrm{k}}) \\ & =-6 x \hat{\mathrm{i}} \\ & \therefore \mathbf{E}(-2,1,0)=-6 \times-2 \hat{\mathrm{i}} \\ & \text { or } \quad \mathrm{E}=12 \hat{\mathrm{i}} \mathrm{V}-\mathrm{m}^{-1} \\ & \therefore \mathrm{E}=12 \mathrm{~V}-\mathrm{m}^{-1} \end{aligned} $

Correct answer: (3)

Easy

Electric Flux

43. In the figure given, $A$ and $B$ are two concentric spherical surfaces. Given that $q _{1}=+1 \mu C, q _{2}=-2 \mu C$ and $q _{3}=+9.854 \mu \mathrm{C}$ and that the flux crossing $B$ is four times that crossing $A$. The value of $Q$ is:

$~$

(1) $8.854 \mu \mathrm{C}$

(2) $17.708 \mu \mathrm{C}$

(3) $26.562 \mu \mathrm{C}$

(4) $35.416 \mu \mathrm{C}$

Show Answer

Correct Answer: (3)

Solution:

The flux crossing spherical surface $A$ is:

$\phi _{\mathrm{A}}=\frac{1}{\varepsilon _{0}} \quad$ (Charge enclosed by A)

or $\quad \phi _{\mathrm{A}}=\frac{1}{\varepsilon _{0}}\left(\mathrm{q} _{1}+\mathrm{q} _{2}+\mathrm{q} _{3}\right)$

or $\quad \phi _{\mathrm{A}}=\frac{1}{\varepsilon _{0}}(1+-2+9.854) \qquad \quad \quad $ (q in $\mu \mathrm{C})$

or $\phi _{\mathrm{A}}=\frac{8.854}{\varepsilon _{0}} \hspace{2cm} \ldots . \ldots {(1)} $

Similarly, $\phi _{\mathrm{B}}=\frac{1}{\varepsilon _{0}}\left(\mathrm{q} _{1}+\mathrm{q} _{2}+\mathrm{q} _{3}+\mathrm{Q}\right)$

$ \begin{aligned} =\frac{1}{\varepsilon _{0}}(8.854+\mathrm{Q}) \hspace{2cm} \ldots . \ldots {(2)} \end{aligned} $

But, $\mathrm{Q} _{\mathrm{B}}=4 \mathrm{QA}$ (given)

$\therefore \frac{1}{\varepsilon _{0}}(8.854+\mathrm{Q})=4 \times \frac{8.854}{\varepsilon _{0}}$

or $\phi=3 \times 8.854$

$=26.562 \mu \mathrm{C}$

Correct answer: (3)

Easy

Electric Flux

44. Point chargers of magnitude $2 Q$ and $-Q$ are located at points $(-a, 0,0)$ and $(4 a, 0,0)$. The ratio of electric flux crossing two spherical surfaces $A$ and $B$ of radii $2 a$ and 8 a respectively, centred at the origin, is:

(1) $1: 4$

(2) $1: 2$

(3) $2: 1$

(4) $1: 1$

Show Answer

Correct Answer: (3)

Solution:

The situation is shown is the figure. Clearly, $2 \mathrm{Q}$ is inside sphere $A$ and both $2 \mathrm{Q}$ and $-\mathrm{Q}$ are inside sphere $\mathrm{B}$.

$$

$$ \begin{align*} & \text{Now,} \phi _{\mathrm{A}}=\frac{1}{\varepsilon _{0}}(2 \mathrm{Q}) \qquad\qquad \ldots . \ldots {(1)}\\ & \phi _{\mathrm{B}}=\frac{1}{\varepsilon _{0}}(2 \mathrm{Q}+-\mathrm{Q}) \\ & =\frac{1}{\varepsilon _{0}}(\mathrm{Q}) \qquad\qquad \ldots . \ldots {(2)}\\ & \therefore \frac{\phi _{\mathrm{A}}}{\phi _{\mathrm{B}}}=2: 1 \end{align*} $$

$~$

Correct answer: (3)

Average

Electric Field due to Charged Sphere

45. At a point which is $20 \mathrm{~cm}$ from the centre of uniformly charged dielectric sphere, of radius $10 \mathrm{~cm}$, the elctric field intensity is $100 \mathrm{Vm}^{-1}$. The elctric field at $3 \mathrm{~cm}$ from the centre of the sphere will be:

(1) $150 \mathrm{Vm}^{-1}$

(2) $125 \mathrm{Vm}^{-1}$

(3) $120 \mathrm{Vm}^{-1}$

(4) zero

Show Answer

Correct Answer: (3)

Solution:

$\mathrm{R}=10 \mathrm{~cm}$

Let $\mathrm{Q}$ be the total charge in the dielectric sphere than

$ \rho=\frac{\mathrm{Q}}{\frac{4}{3} \pi \mathrm{R}^{3}} $

$~$

Now, at $\mathrm{r}=20 \mathrm{~cm} \mathrm{Q}$ acts as point charge at 0 .

$\therefore \mathrm{E} _{\mathrm{A}}=\frac{1}{4 \pi \varepsilon _{0}} \frac{\mathrm{Q}}{\mathrm{r}^{2}}$

or $\quad \frac{\mathrm{Q}}{4 \pi \varepsilon _{0}}=\mathrm{E} _{\mathrm{A}} \mathrm{r}^{2}=100 \times(0.2)^{2}=4$

At $B$, where $r=3 \mathrm{~cm}$, the field is due to the charge inside the sphere of $r=3 \mathrm{~cm}$, which would act as point charge $(q)$ at 0 .

$$ \begin{aligned} \text{Now,} q & =\rho \times \frac{4}{3} \pi r^{3}\\ & =\frac{\mathrm{Q}}{\frac{4}{3} \pi \mathrm{R}^{3}} \times \frac{4}{3} \pi \mathrm{r}^{3} \\ & =\mathrm{Q} \cdot \frac{\mathrm{r}^{3}}{\mathrm{R}^{3}} \\ & =\mathrm{Q} \cdot \frac{(0.03)^{3}}{(0.1)^{3}} \end{aligned} $$

The field at $B$ is:

$$ \begin{aligned} \mathrm{E} _{\mathrm{B}}= & \frac{1}{4 \pi \varepsilon _{0}} \cdot \frac{\mathrm{q}}{\mathrm{r}^{2}} \\ & =\frac{1}{4 \pi \varepsilon _{0}} \times \frac{1}{(0.03)^{2}} \times \mathrm{Q} \cdot \frac{(0.03)^{3}}{(0.1)^{3}} \quad\quad\quad(\mathrm{r}=0.03 \mathrm{~m}) \\ & =\frac{\mathrm{Q}}{4 \pi \varepsilon _{0}} \times \frac{0.03}{0.001} \\ & =4 \times 30 \\ & =120 \mathrm{Vm}^{-1} \end{aligned} $$

Correct answer: (3)

Easy

Capacitance of conductor

46. Taking earth to be a spherical conductor of diameter $12.8 \times 10^{3} \mathrm{~km}$, its electrical capacitance is

(1) $711 \mu \mathrm{F}$

(2) $811 \mu \mathrm{F}$

(3) $611 \mu \mathrm{F}$

(4) $511 \mu \mathrm{F}$

Show Answer

Correct Answer: (1)

Solution:

The capacitance of a spherical conductor is $\mathrm{C}=4 \pi \varepsilon _{0} \mathrm{R}$, where $\mathrm{R}$ is the radius of the spherical conductor. Here, for earth,

$$ \begin{aligned} \mathrm{R}= & \frac{12.8 \times 10^{3}}{2} \mathrm{~km} \\ & =6.4 \times 10^{6} \mathrm{~m} \\ & 4 \pi \varepsilon _{0}=\frac{1}{9 \times 10^{9}} \\ \therefore \mathrm{C}= & \frac{6.4 \times 10^{6}}{9 \times 10^{9}} \\ = & 0.711 \times 10^{3} \mathrm{~F} \\ & =711 \times 10^{-3} \mathrm{~F} \\ & =711 \mu \mathrm{F} \end{aligned} $$

Correct answer: (1)

Average

Electric Force

47. Four charges, equal to $-Q$, are placed at the four corners of a sqare and a charge $q$ is placed at the centre of this square. If the system remains in equilibrium, the value of $q$, in terms of $Q$, is:

(1) $-\frac{\mathrm{Q}}{4}(2 \sqrt{2}+1)$

(2) $\frac{\mathrm{Q}}{4}(1+2 \sqrt{2})$

(3) $-\frac{\mathrm{Q}}{2}(2 \sqrt{2}+1)$

(4) $\frac{\mathrm{Q}}{2}(1+2 \sqrt{2})$

Show Answer

Correct Answer: (2)

Solution:

For equilibrium, the net force should be zero. Taking $-Q$ at $B$, we have

$\mathbf{F} _{\text {BO }}+\mathbf{F} _{\text {BC }}+\mathbf{F} _{\text {BA }}+\mathbf{F} _{\text {BD }}=\mathbf{0}$

$~$

$\left|\mathbf{F} _{\text {Bo }}\right|=\frac{\mathrm{kqQ}}{\left(\frac{\mathrm{d}}{\sqrt{2}}\right)^{2}}$

$\left|\mathbf{F} _{\mathrm{BC}}\right|=\frac{\mathrm{kQ}^{2}}{\mathrm{~d}^{2}} ;\left|\mathbf{F} _{\text {BA }}\right|=\frac{\mathrm{kQ}^{2}}{\mathrm{~d}^{2}}$

$\therefore\left|\mathbf{F} _{\mathrm{BC}}+\mathbf{F} _{\mathrm{BC}}\right|=\sqrt{2} \frac{\mathrm{kQ}^{2}}{\mathrm{~d}^{2}}$, at $45^{0}$ with each other.

Now, $\left|\mathbf{F} _{\mathrm{BD}}\right|=\frac{\mathrm{kQ}^{2}}{(\sqrt{2} \mathrm{~d})^{2}}$, at $45^{0}$, again.

$\therefore\left|\left(\mathbf{F} _{\mathrm{BC}}+\mathbf{F} _{\mathrm{BA}}\right)+\mathbf{F} _{\mathrm{BD}}\right|=\sqrt{2} \frac{\mathrm{kQ}^{2}}{\mathrm{~d}^{2}}+\frac{\mathrm{kQ}^{2}}{2 \mathrm{~d}^{2}}$

For equilibrium,

$ \left|\mathbf{F} _{\mathrm{BO}}\right|=\left|\left(\mathbf{F} _{\mathrm{BC}}+\mathbf{F} _{\mathrm{BA}}\right)+\mathbf{F} _{\mathrm{BD}}\right| $

or $\frac{\mathrm{kqQ}}{\frac{\mathrm{d}^{2}}{2}}=\sqrt{2} \frac{\mathrm{kQ}^{2}}{\mathrm{~d}^{2}}+\frac{\mathrm{kQ}^{2}}{2 \mathrm{~d}^{2}}$

or $\quad q=\frac{Q}{4}(2 \sqrt{2}+1)$

Correct answer: (2)

Easy

Electric Potential

48. A metal sphere $A$ of radius ’ $a$ ’ is charged to some potential $V$. What will be its potential if sphere $A$ is enclosed by spherical conducting shell $B$, of radius, ’ $b$ ’ and the two spheres are connected by a conducting wire?

(1) zero

(2) $\frac{\mathrm{b}}{\mathrm{a}} \mathrm{V}$

(3) $\frac{a}{b} \mathrm{~V}$

(4) $\mathrm{V}$

Show Answer

Correct Answer: (3)

Solution:

Let the charge on A be ’ $q$ ‘, intially.

Then, $\mathrm{V}=\frac{\mathrm{q}}{4 \pi \varepsilon _{0} \mathrm{a}}$

or $\quad \mathrm{q}=\left(4 \pi \varepsilon _{0} \mathrm{a}\right) \mathrm{V}$

When $A$ is placed inside $B$ and connected with wire, the entire charge ’ $q$ ’ flows out on to the surface of B. Now, $A$ is uncharged and it is inside the B, charged to q.

Now, potential on the surface or at any point inside a charged conductor remains the same.

$\therefore$ The potential of $A=$ potential of $B$, now.

$~$

$\mathrm{V} _{\mathrm{A}}=\frac{\mathrm{q}}{4 \pi \varepsilon _{0} \mathrm{~b}}$

or $\quad \mathrm{V} _{\mathrm{A}}=\frac{\left(4 \pi \varepsilon _{0} \mathrm{a}\right) \mathrm{V}}{4 \pi \varepsilon _{0} \mathrm{~b}}=\frac{\mathrm{a}}{\mathrm{b}} \mathrm{V}$

Correct answer: (3)

Difficult

Electric Potential

49. A thin disc of radius $R$ is uniformly charged with $A$ charge. The potential due to this disc, at a point on its axis and distant $R$ from its centre is

(1) $0.707\left(\frac{\mathrm{Q}}{\pi \varepsilon _{0} \mathrm{R}}\right)$

(2) $\frac{\mathrm{Q}}{2 \pi \varepsilon _{0} \mathrm{R}}$

(3) $\frac{\mathrm{Q}}{4 \pi \varepsilon _{0} \sqrt{2 \mathrm{R}}}$

(4) $\frac{\sqrt{2} \mathrm{Q}}{4 \pi \varepsilon _{0} \mathrm{R}}$

Show Answer

Correct Answer: (1)

Solution:

Let $\mathrm{P}$ be a point on the axis of the disc, at distance $\mathrm{z}$ from its centre.

Assuming that the disc be divided large number of elementary rings, each ring will produced a potential of $d V$ at $P$. Intergration over all the $d V$ s will give $V$ at $P$.

The charge on one such elementary ring, as shown in the figure is:

$~$

$\mathrm{dq}=(2 \pi \mathrm{rdr}) \sigma \quad \quad \quad \quad \left(\sigma=\frac{\mathrm{Q}}{\pi \mathrm{R}^{2}}\right)$

So, $\mathrm{dv}=\frac{1}{4 \pi \varepsilon _{0}} \frac{(\mathrm{dq})}{\mathrm{d}}$

$ =\frac{1}{4 \pi \varepsilon _{0}} \frac{2 \pi \mathrm{rdr} \sigma}{\sqrt{\mathrm{z}^{2}+\mathrm{r}^{2}}} $

$\therefore$ Potential at $\mathrm{P}$ due to the entire disc is:

$$ \begin{aligned} \mathrm{V}= & \frac{\sigma \pi}{4 \pi \varepsilon _{0}} \int _{0}^{\mathrm{R}} \frac{2 \mathrm{rdr}}{\sqrt{\mathrm{r}^{2}+\mathrm{z}^{2}}} \\ & =\frac{2 \sigma \pi}{4 \pi \varepsilon _{0}}\left[\sqrt{\mathrm{r}^{2}+\mathrm{z}^{2}}\right] _{0}^{\mathrm{R}} \\ & =\frac{2 \sigma \pi}{4 \pi \varepsilon _{0}}\left[\sqrt{\mathrm{r}^{2}+\mathrm{z}^{2}}-\sqrt{0+\mathrm{z}^{2}}\right] \\ & =\frac{2 \sigma \pi}{4 \pi \varepsilon _{0}}\left[\sqrt{\mathrm{r}^{2}+\mathrm{z}^{2}}-\mathrm{z}\right] \\ & =\frac{2 \mathrm{Q}}{4 \pi \varepsilon _{0} \mathrm{R}^{2}}\left[\sqrt{\mathrm{r}^{2}+\mathrm{z}^{2}}-\mathrm{z}\right] \end{aligned} $$

Now, at $\mathrm{z}=\mathrm{R}$,

$$ \begin{array}{r} \mathrm{V}=\frac{2 \mathrm{Q}}{4 \pi \varepsilon _{0} \mathrm{R}^{2}}[\sqrt{2} \mathrm{R}-\mathrm{R}] \\ =\frac{\mathrm{Q}}{4 \pi \varepsilon _{0} \mathrm{R}}[\sqrt{2}-1] \\ \quad=0.707\left(\frac{\mathrm{Q}}{\pi \varepsilon _{0} \mathrm{R}}\right) \end{array} $$

Correct answer: (1)

Difficult

Electric Field

50. A simple pendulum, of length ’ $l$ ‘, has a bob of mass ’ $m$ ‘. A charge ’ $q$ ’ is given to the bob and it is made to execute simple harmonic oscillations parallel to the field lines of a horizontal uniform electric field $\mathrm{z}$. If ’ $\mathrm{g}$ ’ is the acceleration due to gravity, then the time period of oscillation of the pendulum is

(1) $ 2 \pi \sqrt{\frac{\ell}{\sqrt{g^{2}+\left(\frac{q E}{m}\right)^{2}}}}$

(2) $ T=2 \pi \sqrt{\frac{\mathrm{m} \ell}{\sqrt{g^{2}+\left(\frac{\mathrm{qE}}{\mathrm{m}}\right)^{2}}}}$

(3) $\mathrm{T}=2 \pi \sqrt{\frac{\ell}{\sqrt{(\mathrm{mg})^{2}+(\mathrm{qE})^{2}}}}$

(4) $\mathrm{T}=2 \pi \sqrt{\frac{\ell}{\sqrt{\mathrm{q}^{2}+(\mathrm{qE})^{2}}}} $

Show Answer

Correct Answer: (1)

Solution:

There are two forces on the bob, as shown in the figure. (i) $\mathrm{mg}$ downwords (ii) $\mathrm{qE}$, horizontally.

So the resultant force on the bob is $\mathrm{F}=\sqrt{(\mathrm{mg})^{2}+(\mathrm{qE})^{2}}$

In equilibrium position of the bob, this for $F$ will act along the length of the pendulum. So the pendulum will rest at certain angle from the vertical.

Now, suppose the pendulum is disturbed through a small angle $\theta$, by moving the bob in the plane of the electric field, as shown.

The restoring force developed on the bob will be:

$~$

$\mathrm{F}^{\prime}=\mathrm{F} \sin \theta$, directed opposite to $\mathrm{S}$.

$F$ ’ acts on the bob. So when the bob is released, $F$ ’ will produce and acceleration, ’ $a$ ‘, such that $F$ ’ $=$ ma.

$\therefore \mathrm{ma}=-\mathrm{Fsin} \theta \quad$ (-ve sign as the force is opposite to $\mathrm{S}$ )

or $\quad \frac{\mathrm{d}^{2} \mathrm{~S}}{\mathrm{dt}^{2}}=-\frac{\mathrm{F}}{\mathrm{m}} \cdot \theta$

or $\quad \frac{\mathrm{d}^{2} \mathrm{~S}}{\mathrm{dt}^{2}}=-\sqrt{\frac{(\mathrm{mg})^{2}+(\mathrm{qE})^{2}}{\mathrm{~m}}} \cdot\left(\frac{\mathrm{S}}{\ell}\right)$

or $\quad \frac{\mathrm{d}^{2} \mathrm{~S}}{\mathrm{dt}^{2}}=-\left(\frac{\sqrt{\mathrm{g}^{2}+\left(\frac{\mathrm{qE}}{\mathrm{m}}\right)^{2}}}{\ell}\right) \mathrm{S}$

This is of the form, $\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dt}^{2}}=-\omega^{2} \mathrm{y}$ for S.H.M.

$\therefore \omega^{2}=\sqrt{\frac{\mathrm{g}^{2}+\left(\frac{\mathrm{qE}}{\mathrm{m}}\right)^{2}}{\ell}}$ or $\omega=\sqrt{\frac{\sqrt{\mathrm{g}^{2}+\left(\frac{\mathrm{qE}}{\mathrm{m}}\right)^{2}}}{\ell}}$

Now, $\mathrm{T}=\frac{2 \pi}{\omega}$

or $\quad T=2 \pi \sqrt{\frac{\ell}{\sqrt{g^{2}+\left(\frac{q E}{m}\right)^{2}}}}$

Correct answer: (1)

Average

Motion in Electric Field

51. A small copper ball, of density $8.6 \mathrm{gcm}^{-3}$ and diameter $4 \mathrm{~cm}$, is immersed in oil (an insulator) of density $0.8 \mathrm{gcm}^{-3}$. This copper ball, of charge ’ $q$ ‘, just remain suspented in the oil, under the effect of an electric field of intensity $2600 \mathrm{Vm}^{-1}$ acting in the upward direction. The value of $q$ is: $\left(\right.$ Take $g=10 \mathrm{~ms}^{-2}$ )

(1) $47.54 \mathrm{mC}$

(2) $43.12 \mathrm{mC}$

(3) $-47.54 \mu \mathrm{C}$

(4) $43.12 \mu \mathrm{C}$

Show Answer

Correct Answer: (2)

Solution:

The ball will remain suspended in oil when the force due to electric field balances the apparent weight of the ball in the oil.

$\therefore \mathrm{qE}=$ Apparent weight

or $\quad \mathrm{qE}=\frac{4}{3} \pi \mathrm{r}^{3}(\rho-\sigma) \mathrm{g}$

or $\quad q=\frac{\frac{4}{3} \pi r^{3}(\rho-\sigma) g}{E}$

Here, Density of copper $(\rho)=8.6 \mathrm{gcm}^{-3}=8600 \mathrm{kgm}^{-3}$

Density of oil $(\sigma)=0.8 \mathrm{gcm}^{-3}=800 \mathrm{kgm}^{-3}$

$\mathrm{r}=\frac{14 \mathrm{~cm}}{2}=7 \mathrm{~cm}=7 \times 10^{-2} \mathrm{~m}$

$\mathrm{E}=2600 \mathrm{Vm}^{-1}$

$\mathrm{g}=10 \mathrm{~ms}^{-2}$

$\therefore \mathrm{q}=\frac{4}{3} \times \frac{22}{7} \times \frac{\left(7 \times 10^{-2}\right)(8600-800) \times 10}{2600}$

$=88 \times 49 \times 10^{-5}$

$=4312 \times 10^{-5} \mathrm{C}$

$=43.12 \mathrm{mC}$

Correct answer: (2)

Average

Principle of Superposition of Charges

52. Point charges of value $4 \mu \mathrm{C}$ and $-9 \mu \mathrm{C}$ are placed at certain distance apart in air. A third charge ’ $q$ ’ is placed, on the line joining these two charges, and at a suitable position, such that all the three charges remain in equilibrium. The value of ’ $q$ ’ is

(1) $+36 \mu \mathrm{C}$

(2) $+1.44 \mu \mathrm{C}$

(3) $-36 \mu \mathrm{C}$

(4) $-1.44 \mu \mathrm{C}$

Show Answer

Correct Answer: (3)

Solution:

For equilibrium, there should be two equal and opposite forces on each charge. This condition is statified, here, only if the third charge ’ $q$ ’ is placed on the line joining the first two, outside the two, and nearer to the smaller charge. This is shown diagramtically as.

$~$

And the third charge should be negative.

For charge $\mathrm{q}$ to be is equibrium,

$$ \begin{aligned} & \quad \mathrm{F} _{31}=\mathrm{F} _{32} \\ & \text { or } \quad \frac{\mathrm{Kq} \times 4}{x^{2}}=\frac{\mathrm{Kq} \times 9}{(x+\mathrm{d})^{2}} \\ & \text { or } \quad \frac{2}{x}=\frac{3}{x+\mathrm{d}} \\ & \therefore \quad x=2 \mathrm{~d} \end{aligned} $$

For charge $+4 \mu \mathrm{C}$ to be in equilibrium,

$$ \begin{aligned} & \quad F _{12}=F _{13} \\ & \text { or } \quad \frac{K \times 4 \times 9}{d^{2}}=\frac{K \times 4 \times q}{x^{2}} \\ & \text { or } \quad \frac{9}{d^{2}}=\frac{q}{(2 d)^{2}} \\ & \therefore q=36 \mu \mathrm{C} \quad \text { (Negative) } \\ & \text { or } q=-36 \mu \mathrm{C} \end{aligned} $$

Correct answer: (3)

Average

Motion of charge in uniform Electric Field

53. A particle of mass ’ $m$ ’ and carrying a charge ’ $q$ ’ is kept at rest at point $A$, is a uniform electric field of intensity E. At time instant $t=0$, the particle released from $A$. At $t=t$, it hits a block at $B(A B=x)$ elastically, and charges its direction of motion by $\mathbf{9 0}^{\circ}$, instantly. Following a parabolic path BCD, it reaches point $D$, as shown in the figure, at time $t=2 t$. The speed of the particle at $D$ is:

$~$

(1) $\frac{1}{2} \mathrm{qE} x^{2}$

(2) $\left(\frac{2 E x^{2}}{m}\right)^{1 / 2}$

(3) $2 \sqrt{\frac{E x q}{m}}$

(4) $\left(\frac{\mathrm{qE} x}{\mathrm{~m}}\right)^{1 / 2}$

Show Answer

Correct Answer: (3)

Solution:

The electric field will produce a constant acceleration of $a=\frac{q E}{m}$ along the field direction.

The velocity attained by $q$ at $B$ is:

$ \mathrm{v}^{2}=0+2 \frac{\mathrm{qE}}{\mathrm{m}} x $

or $\quad v=\sqrt{\frac{2 \mathrm{qE}}{\mathrm{m}}} x$

After collision, this $v$ become perpendicular to $E$ and hence remains constant (along $-\hat{\mathbf{j}}$ ). But, the

acceleration remains the same in magnitude and direction (along $\hat{\mathrm{i}}$ ). Now, along E-field, the particle again will cover $x$ distance in the next $t$ second after collision, as it reaches the point $\mathrm{D}$, following the parabolic path.

Hence, its velocity, will again increases to $v$ (along E-field) where $v=\sqrt{\frac{2 \mathrm{qE}}{\mathrm{m}}} x$

Therefore, at $D$ it will have $\mathbf{v} _{\mathrm{P}}=v \hat{\mathrm{i}}-\mathrm{v} \hat{\mathrm{j}}$

$\therefore$ Speed at ’ $\mathrm{D}$ ’ will be

$$ \begin{aligned} & \mathrm{v} _{\mathrm{D}}=\sqrt{\mathrm{v}^{2}+\mathrm{v}^{2}} \\ & =\sqrt{2 \mathrm{v}^{2}} \\ & =\sqrt{2 \times \frac{2 \mathrm{qE}}{\mathrm{m}}} x \\ & =2 \sqrt{\frac{\mathrm{Exq}}{\mathrm{m}}} \end{aligned} $$

Correct answer: (3)

Difficult

Electric field

54. Two co-axial circular rings of radius $R=10 \mathrm{~cm}$ each are uniformly charged with $Q=\sqrt{3} \mathrm{nC}$ each. The distance between their centres is such that the centre of one ring, produced by the other is maximum. The magnitude of the net electric field at the centre of each ring is

$~$

(1) $600 \mathrm{NC}^{-1}$

(2) $900 \sqrt{3} \mathrm{NC}^{-1}$

(3) $(6+9 \sqrt{3}) \times 10^{2} \mathrm{NC}^{-1}$

(4) zero

Show Answer

Correct Answer: (1)

Solution:

When the field is maximum, the distance $x=\frac{\mathrm{R}}{\sqrt{2}}$, according to the formula

$$ \mathrm{E}=\frac{\mathrm{KQ} x}{\sqrt{\mathrm{R}^{2}+x^{2}}} \quad(x \text { is distance along the axis }) $$

The field due to the charge on the ring at its own centre is always zero.

$\therefore$ The net field at the centre of one is

$$ \mathrm{E} _{\text {net }}=0+\frac{\mathrm{KQ} x}{\sqrt{\mathrm{R}^{2}+x^{2}}}, \text { where } x=\frac{\mathrm{R}}{\sqrt{2}} $$

$$ \text{Here,} \mathrm{R}=10 \mathrm{~cm}=10 \times 10^{-2} \mathrm{~m}$$

$$ \begin{aligned} & \frac{\mathrm{R}}{\sqrt{2}}=\frac{10}{\sqrt{2}} \mathrm{~cm}=5 \sqrt{2} \mathrm{~cm}=5 \sqrt{2} \times 10^{-2} \mathrm{~m} \\ & \mathrm{Q}=\sqrt{3} \mathrm{nC}=\sqrt{3} \times 10^{-9} \mathrm{C} \\ & \mathrm{K}=9 \times 10^{9} \text { SI Units } \end{aligned} $$

$$ \begin{aligned} \mathrm{E} _{\text {net }}= & \frac{9 \times 10^{9} \times \sqrt{3} \times 10^{-9} \times 5 \sqrt{2} \times 10^{-2}}{\left[\left(10 \times 10^{-2}\right)^{2}+\left(5 \sqrt{2} \times 10^{-2}\right)^{2}\right]^{3 / 2}} \\ = & \frac{9 \times \sqrt{3} \times 5 \sqrt{2} \times 10^{-2}}{(100+50)^{3 / 2} \times 10^{-6}} \end{aligned} $$

$$ \begin{aligned} & =\frac{9 \times \sqrt{3} \times 5 \sqrt{2} \times 10^{-2}}{\left(5^{2} \times 6\right)^{3 / 2} \times 10^{-6}} \\ & =\frac{9 \times \sqrt{3} \times 5 \sqrt{2} \times 10^{4}}{5^{3} \times 6 \times \sqrt{3} \times \sqrt{2}} \\ \\ & =600 \mathrm{NC}^{-1} \end{aligned} $$

Correct answer: (1)

Easy

Continuous Charge Distribution

55. A conducting sphere $A$, of radius $10 \mathrm{~cm}$ is charged to $10 \mu \mathrm{C}$. Another uncharged conducting sphere $B$, of radius $20 \mathrm{~cm}$ is brought is to contact with $A$ and after that the spheres are separated to large distance. The surface density of charge on the spheres, now, will be in the ratio:

(1) $1: 4$

(2) $4: 2$

(3) $1: 2$

(4) $2: 1$

Show Answer

Correct Answer: (4)

Solution:

When the spheres are brought to contact, they share the total charge in such a way that the potential of both are same.

i.e. $V _{A}=V _{B} \quad$ (after they are brought in to contact).

If $Q _{A}$ and $Q _{B}$ are the final charges on the spheres, then

$~$

$\mathrm{V} _{\mathrm{A}}=\frac{\mathrm{KQ} _{\mathrm{A}}}{\mathrm{r} _{\mathrm{A}}}$ and $\mathrm{V} _{\mathrm{B}}=\frac{\mathrm{KQ} _{\mathrm{B}}}{\mathrm{r} _{\mathrm{B}}}$

Since $V _{A}=V _{B}$,

$ \Rightarrow \frac{Q _{A}}{r _{A}}=\frac{Q _{B}}{r _{B}} $

or $\quad \frac{4 \pi \mathrm{r} _{\mathrm{A}}^{2} \sigma _{\mathrm{A}}}{\mathrm{r} _{\mathrm{A}}}=\frac{4 \pi \mathrm{r} _{\mathrm{B}}^{2} \sigma _{\mathrm{B}}}{\mathrm{r} _{\mathrm{B}}}$

$\therefore \frac{\sigma _{\mathrm{A}}}{\sigma _{\mathrm{B}}}=\frac{\mathrm{r} _{\mathrm{B}}}{\mathrm{r} _{\mathrm{A}}}=2: 1$

Correct answer: (4)

Easy

Electric Dipole

56. A dipole consisting of two point charges of $-2 \mu \mathrm{C}$ and $+2 \mu \mathrm{C}$ located at points $A(1,2,2) \mathrm{cm}$ and $B(2,1,4)$ cm respectively, is in a uniform electric field of $E=(3 \hat{i}+\hat{j}+2 \hat{k}) \times 10^{5} \mathrm{Vm}^{-1}$. The magnitude of torque experienced by the dipole is

(1) $2.772 \times 10^{-2} \mathrm{Nm}$

(2) $1.386 \mathrm{Nm}$

(3) $\quad 2.772 \mathrm{Nm}$

(4) $1.386 \times 10^{-2} \mathrm{Nm}$

Show Answer

Correct Answer: (4)

Solution:

Let the dipole moment be $\overrightarrow{\mathrm{P}}$

$$ \mathbf{P}=(2 \mathbf{a}) \mathrm{q} \quad\quad\quad (2 \mathbf{a}=\text { distance vector from }-\mathrm{q} \text { to }+\mathrm{q}) $$

Now, $2 \mathbf{a}=\mathbf{r} _{B}-\mathbf{r} _{\mathrm{A}}$

$ \begin{aligned} = & (2 \hat{i}+\hat{j}+4 \hat{k})-(\hat{i}+2 \hat{j}+4 \hat{k}) \\ & =(\hat{i}-\hat{j}+2 \hat{k}) \mathrm{cm} \\ & =(\hat{i}-\hat{j}+2 \hat{k}) \times 10^{-2} \mathrm{~m} \\ q=2 & \mu C=2 \times 10^{-6} \mathrm{C} \end{aligned} $

$\therefore \mathbf{P}=(2 \hat{\mathrm{i}}+2 \hat{\mathbf{j}}+4 \hat{\mathrm{k}}) \times 10^{-8} \mathrm{~cm}$

$\mathbf{E}=(3 \hat{\mathrm{i}}+\hat{\mathrm{j}}+2 \hat{\mathrm{k}}) \times 10^{6} \mathrm{Vm}^{-1}$

$\therefore \boldsymbol{\tau}=\mathbf{P} \times \mathbf{E}=10^{-3} \times\left|\begin{array}{ccc}\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \ 2 & -2 & 4 \ 3 & 1 & 2\end{array}\right|=[\hat{\mathrm{i}}(-4-4)+\hat{\mathrm{j}}(12-4)+\hat{\mathrm{k}}(2+6)] \times 10^{-3}$

$=[-8 \hat{\mathrm{i}}+8 \hat{\mathrm{j}}+8 \hat{\mathrm{k}}] \times 10^{-3}$

$\therefore \boldsymbol{\tau}=|\boldsymbol{\tau}|=8 \sqrt{3} \times 10^{-3} \mathrm{Nm}$

$=1.386 \times 10^{-2} \mathrm{Nm}$

Correct answer: (4)

Average

Principle of superposition of Charges

57. An infinite number of point charges, each of magnitude $q=50 \mu \mathrm{C}$, are placed along the $x-$ axis, at $x=1 \mathrm{~m}, x=2 \mathrm{~m}, x=4 \mathrm{~m}, x=8 \mathrm{~m}, x=16 \mathrm{~m}$ and so 0 . Consecutive charges have opposite sign, with $q$ at $x=1 \mathrm{~m}$ being positive. The net electric field at the origin 0 has a magnitude of

(1) $6.0 \times 10^{5} \mathrm{NC}^{-1}$

(2) $6.0 \times 10^{9} \mathrm{NC}^{-1}$

(3) $3.6 \times 10^{5} \mathrm{NC}^{-1}$

(4) $2.7 \times 10^{5} \mathrm{NC}^{-1}$

Show Answer

Correct Answer: (3)

Solution:

$$ \begin{aligned} & \mathrm{E}=\frac{1}{4 \pi \varepsilon _{0}}\left[\frac{\mathrm{q}}{1^{2}}-\frac{\mathrm{q}}{2^{2}}+\frac{\mathrm{q}}{4^{2}}-\frac{\mathrm{q}}{8^{2}}+\frac{\mathrm{q}}{16^{2}} \ldots \ldots .\right] \\ & =\frac{\mathrm{q}}{4 \pi \varepsilon _{0}}\left[\frac{1}{1}-\frac{1}{4}+\frac{1}{16}-\frac{1}{64}+\frac{1}{265} \ldots . . .\right] \\ & =\frac{\mathrm{q}}{4 \pi \varepsilon _{0}}\left[\frac{1}{1-\left(\frac{-1}{4}\right)}\right] \quad\quad\quad \left(\because \mathrm{S}=\frac{\mathrm{a}}{1-\mathrm{r}} \text { for sum of infnite terms of a G.P. }\right) \\ & =9 \times 10^{9} \times 50 \times 10^{-6} \times \frac{4}{5} \\ & =3.6 \times 10^{5} \mathrm{NC}^{-1} \end{aligned} $$

Correct answer: (3)

Average

Electric Field Lines

58. The figure shows a few electric field lines for a system of two charges $q _{1}$ and $q _{2}$, fixed at two different points on the $\boldsymbol{x}$-axis. These field lines suggest that

$~$

(1) $\left|q _{1}\right|>\left|q _{2}\right|$ and at a finite distance, to the right of $q _{2}$, the electric field is zero.

(2) $\left|q _{1}\right|<\left|q _{2}\right|$ and at a finite distance, to the left of $q _{1}$, the electric field is zero.

(3) $\left|q _{1}\right|=\left|q _{2}\right|$ and there no point at which the electric field is zero.

(4) $\left|q _{1}\right|>\left|q _{2}\right|$ and at a finite point to the left of $q _{1}$, the electric field is zero.

Show Answer

Correct Answer: (1)

Solution:

Here, $\mathrm{q} _{1}$ is positive and $\mathrm{q} _{2}$ is negative as field lines originate from positive change and terminate at negative change. Also, more lines are drawn fro $q _{1}$, indicating that its magnitude is larger as compared to $q _{2}$ in the diagram.

$$ \therefore\left|\mathrm{q} _{1}\right|>\left|\mathrm{q} _{2}\right| $$

Now, for $+q _{1}$ and $-q _{2}$, the null point, where $E=0$, is obtained on the line joining the two charges, outside than and nearer to the smaller charge.

Hence field is zero at some distance to the right of $\mathrm{q} _{2}$.

Easy

Surface Charge Density

59. A uniformly charged thin spherical shell, made of a co-conducting material, of radius $R$, carries a uniform surface charge density of $\sigma$. It is made of two hemispherical shell,s held together by pressing then with force $F$, as shown in the figure. $F$ is proportional to:

$~$

(1) $\frac{1}{\varepsilon _{0}} \sigma^{2} R^{2}$

(2) $\frac{1}{\varepsilon _{0}} \sigma^{2} \mathrm{R}$

(3) $\frac{1}{\varepsilon _{0}} \frac{\sigma^{2}}{R}$

(4) $\frac{1}{\varepsilon _{0}} \frac{\sigma^{2}}{\mathrm{R}^{2}}$

Show Answer

Solution:

According to the principles of homofensity of dimensions, the correct option is the one which has dimensions same as that of force i.e. $\mathrm{M}^{1} \mathrm{~L}^{1} \mathrm{~T}^{-2}$.

Now, $\left[\frac{1}{\varepsilon _{0}}\right]=\left[\frac{\mathrm{F} \times \mathrm{r}^{2}}{\mathrm{q}^{2}}\right]=\frac{\mathrm{M} _{1} \mathrm{~L} _{1} \mathrm{~T}^{-2} \times \mathrm{L}^{2}}{\mathrm{~A}^{2} \mathrm{~T}^{2}}=\mathrm{M}^{1} \mathrm{~L}^{3} \mathrm{~T}^{-4} \mathrm{~A}^{-2}$

$\left[\sigma^{2}\right]=\left[\frac{\mathrm{q}}{\mathrm{S}}\right]^{2}=\left[\frac{\mathrm{A}^{1} \mathrm{~T}^{1}}{\mathrm{~L}^{2}}\right]^{2}=\mathrm{A}^{2} \mathrm{~T}^{2} \mathrm{~L}^{-4}$

$[\mathrm{R}]=\mathrm{L}^{\prime}$

1. $\left[\frac{1}{\varepsilon _{0}} \sigma^{2} R^{2}\right]=M^{1} L^{3} T^{-4} A^{-2} \times A^{2} T^{2} L^{-4} \times L^{2}$

$ \qquad =\mathrm{ML}^{1} \mathrm{~T}^{-2}=[\mathrm{F}] $

2. $\left[\frac{1}{\varepsilon _{0}} \sigma^{2} \mathrm{R}\right]=\mathrm{M}^{1} \mathrm{~L}^{3} \mathrm{~T}^{-4} \mathrm{~A}^{-2} \times \mathrm{A}^{2} \mathrm{~T}^{2} \mathrm{~L}^{-4} \times \mathrm{L}^{1}$

$ \qquad =\mathrm{ML}^{2} \mathrm{~T}^{-2} \neq[\mathrm{F}]$

3. $\left[\frac{1}{\varepsilon _{0}} \frac{\sigma^{2}}{R}\right]=M^{1} L^{3} T^{-4} \mathrm{~A}^{-2} \times \mathrm{A}^{2} \mathrm{~T}^{2} \mathrm{~L}^{-4} \times \mathrm{L}^{1}$

$ \qquad =\mathrm{M}^{1} \mathrm{~L}^{-2} \mathrm{~T}^{-2} \neq[\mathrm{F}]$

4. $\left[\frac{1}{\varepsilon _{0}} \frac{\sigma^{2}}{\mathrm{R}^{2}}\right]=\mathrm{M}^{1} \mathrm{~L}^{3} \mathrm{~T}^{-4} \mathrm{~A}^{-2} \times \mathrm{A}^{2} \mathrm{~T}^{2} \mathrm{~L}^{-4} \times \mathrm{L}^{-2}$

$ \qquad =\mathrm{M}^{1} \mathrm{~L}^{-3} \mathrm{~T}^{-2} \neq[\mathrm{F}] $

Correct answer: (1)

Difficult

Electric Field

60. A thin semicircular ring of radius ’ $r$ ’ has a positive charge ’ $q$ ’ distributed uniformly over it. The net field $\mathbf{E}$ at its centre is:

$~$

(1) $\frac{\mathrm{q}}{2 \pi^{2} \varepsilon _{0} \mathrm{r}^{2}} \hat{\mathrm{j}}$

(2) $\frac{q}{4 \pi^{2} \varepsilon _{0} r^{2}} \hat{j}$

(3) $\frac{-q}{4 \pi^{2} \varepsilon _{0} r^{2}} \hat{j}$

(4) $\frac{-q}{2 \pi^{2} \varepsilon _{0} r^{2}} \hat{j}$

Show Answer

Correct Answer: (4)

Solution:

Consider an element of length $\mathrm{d} \ell$ at angle $\theta$.

The charge on this element on this element is $\mathrm{dq}=\lambda \mathrm{d} \ell$

$$ \begin{aligned} & \mathrm{d} \ell=\mathrm{rd} \theta \\ & \therefore \mathrm{dq}=\lambda \mathrm{rd} \theta \end{aligned} $$

$~$

Now the field $\mathrm{dE}$ at 0 due to this element is:

$$ \begin{aligned} & \mathrm{dE}=\frac{1}{4 \pi \varepsilon _{0}} \cdot \frac{\lambda \mathrm{rd} \theta}{\mathrm{r}^{2}} \\ & =\frac{1}{4 \pi \varepsilon _{0}} \cdot\left(\frac{\lambda}{\mathrm{r}}\right) \mathrm{d} \theta, \text { as shown in the figure. } \end{aligned} $$

Resolving $\mathrm{dE}$ is to two components, the cos components vanish and sin components add-up along $(-\hat{\mathrm{j}})$ direction.

$$ \begin{aligned} \therefore \mathrm{E} & =\int \mathrm{dE} \sin \theta \qquad \qquad \qquad (\text{along} -\hat{\mathrm{j}})\\ & =\int \frac{1}{4 \pi \varepsilon _{0}} \frac{\lambda}{\mathrm{r}} \cdot \mathrm{d} \theta \sin \theta \\ & =\int \frac{\lambda}{4 \pi \varepsilon _{0} \mathrm{r}} \int _{0^{0}}^{180^{0}} \sin \theta \mathrm{d} \theta=\frac{\lambda}{4 \pi \varepsilon _{0} \mathrm{r}}[-\cos \theta] _{0^{0}}^{180^{0}} \\ & =\frac{\lambda}{4 \pi \varepsilon _{0} \mathrm{r}}[\cos \theta] _{0^{0}}^{180^{0}}=\frac{\lambda}{4 \pi \varepsilon _{0} \mathrm{r}}(1-(-1)) \\ & =\frac{2 \lambda}{4 \pi \varepsilon _{0} \mathrm{r}}=\frac{\mathrm{q}}{2 \pi \varepsilon _{0} \mathrm{r}^{2}} \quad \quad \quad \left(\lambda=\frac{\mathrm{q}}{\pi \mathrm{r}}\right) \\ \text { or } \quad \mathbf{E} & =-\frac{\mathrm{q}}{2 \pi \varepsilon _{0} \mathrm{r}^{2}}(\hat{\mathrm{j}}) \end{aligned} $$

Correct answer: (4)

Difficult

Continuous Charge Distribution

61. A spherically symmetric charge distribution is set-up with the charge density varying according to the expression: $\rho _{(\mathrm{r})}=\rho _{(0)}\left(\frac{5}{4}-\frac{r}{R}\right)$, up to $r=R$ and for $r>R, \rho _{(r)}=0$. Here $r$ is the distance from the centre $O$ of the spherical distribution. The electric field intensity at a distance $r(r<R)$, from $O$ is given by:

(1) $\frac{\rho _{(0)} \mathrm{r}}{3 \varepsilon _{0}}\left(\frac{5}{4}-\frac{\mathrm{r}}{\mathrm{R}}\right)$

(2) $\frac{\rho _{(0)} 4 \pi \mathrm{r}}{3 \varepsilon _{0}}\left(\frac{5}{4}-\frac{\mathrm{r}}{\mathrm{R}}\right)$

(3) $\frac{\rho _{(0)} \mathrm{r}}{4 \varepsilon _{0}}\left(\frac{5}{3}-\frac{\mathrm{r}}{\mathrm{R}}\right)$

(4) $\frac{4 \rho _{(0)} \mathrm{r}}{3 \varepsilon _{0}}\left(\frac{5}{3}-\frac{\mathrm{r}}{\mathrm{R}}\right)$

Show Answer

Correct Answer: (3)

Solution:

Consider a thin elementary of radius $r$ and thickness dr. The charge of this shell is:

$$ \begin{aligned} & \mathrm{dq}=\rho _{(\mathrm{r})} 4 \pi \mathrm{r}^{2} \mathrm{dr} \\ & =\rho _{(0)}\left(\frac{5}{4}-\frac{r}{R}\right) 4 \pi r^{2} d r \\ & =\frac{5}{4} \rho _{(0)} 4 \pi r^{2} d r-\frac{\rho _{(0)}}{R} 4 \pi r^{3} d r \end{aligned} $$

$~$

$\therefore$ Total charge in the spherical region of radius $r$ is:

$$ \begin{gathered} \mathrm{Q}=\int _{\mathrm{r}=0}^{\mathrm{r}=\mathrm{r}} \mathrm{dQ}=\frac{5}{4} \rho _{(0)} 4 \pi \frac{\mathrm{r}^{3}}{3}-\frac{\rho _{(0)}}{\mathrm{R}} 4 \pi \frac{\mathrm{r}^{4}}{4} \\ =\pi \rho _{(0)} \mathrm{r}^{3}\left[\frac{5}{3}-\frac{\mathrm{r}}{\mathrm{R}}\right] \end{gathered} $$

Now, $E=\frac{K Q}{\mathrm{r}^{2}}=\frac{1}{4 \pi \varepsilon _{0}} \frac{\theta}{\mathrm{r}^{2}}$

$$ \begin{aligned} & =\frac{1}{4 \pi \varepsilon _{0} \mathrm{r}^{2}} \times \pi \rho _{(0)} \mathrm{r}^{3}\left[\frac{5}{3}-\frac{\mathrm{r}}{\mathrm{R}}\right] \\ & =\frac{\rho _{(0)} \mathrm{r}}{4 \varepsilon _{0}}\left[\frac{5}{3}-\frac{\mathrm{r}}{\mathrm{R}}\right] \end{aligned} $$

Correct answer: (3)

Easy

Electric Field and Potential

62. Three point charges of $2 q,-q$ and $-q$ are located at the vertices of an equilateral triangle. At the circumcentre of the triangle.

(1) The field is non-zero and the potential is zero.

(2) The field is zero and the potential is non-zero.

(3) Both field and potential are non-zero.

(4) Both field and potential are zero.

Show Answer

Correct Answer: (1)

Solution:

As shown in the figure, the resultant of fields $\mathrm{E} _{1}, \mathrm{E} _{2}$ and $\mathrm{E} _{3}$ can’t be zero.

But the potential is:

$\mathrm{V}=\mathrm{V} _{1}+\mathrm{V} _{2}+\mathrm{V} _{3}=\frac{\mathrm{k}(2 \mathrm{q})}{\mathrm{d}}-\frac{\mathrm{kq}}{\mathrm{d}}-\frac{\mathrm{kq}}{\mathrm{d}}=0$

$~$

Correct answer: (1)

Easy

Electrostatic Potential Energy

63. Three point charges of $1 \mathrm{C}, 2 \mathrm{C}$ and $3 \mathrm{C}$ are placed at the corners of an equitateral triangle of side $3 \mathrm{~m}$. The work done is bringing these charges closer by $1 \mathrm{~m}$, towards each other is:

(1) $6.6 \times 10^{10} \mathrm{~J}$

(2) $1.65 \times 10^{10} \mathrm{~J}$

(3) $1.8 \times 10^{10} \mathrm{~J}$

(4) $3.3 \times 10^{10} \mathrm{~J}$

Show Answer

Correct Answer: (2)

Solution:

The intial and final arrangements are as shown in the diagram.

$~$

$U _{i}=\frac{k q _{1} q _{2}}{d}+\frac{k q _{2} q _{3}}{d}+\frac{k q _{3} q _{1}}{d}$

$ \begin{aligned} & =\frac{9 \times 10^{9}}{3}[1 \times 2+2 \times 3+3 \times 1] \\ & =3 \times 10^{9} \times 11 \mathrm{~J} \end{aligned} $

$\mathrm{U} _{\mathrm{f}}=\frac{9 \times 10^{9}}{2} \times 11=4.5 \times 10^{9} \times 11 \mathrm{~J}$

$\therefore \mathrm{W}=\mathrm{U} _{\mathrm{f}}-\mathrm{U} _{\mathrm{i}}=1.5 \times 10^{9} \times 11$

$ =1.65 \times 10^{10} \mathrm{~J} $

Correct answer: (2)

Average

Electric Potential

64. For the rectangle $\mathrm{ABCD}$ in the figure, $\mathrm{E}$ and $\mathrm{F}$ are mid points of side $A B$ and side $D C$ respectively. If $q=+10 \mathrm{nC}$, then the potential difference between $\mathrm{E}$ and $\mathrm{E}$ is:

$A B=C D=6 m$

$\mathrm{CB}=\mathbf{A D}=\mathbf{4 ~ m}$

$~$

(1) $1.2 \mathrm{~V}$

(2) $12 \mathrm{~V}$

(3) $15 \mathrm{~V}$

(4) $18 \mathrm{~V}$

Show Answer

Correct Answer: (2)

Solution:

The potential at point $\mathrm{E}$ is

$~$

$ \begin{aligned} \mathrm{V} _{\mathrm{E}}= & \mathrm{k}\left(\frac{\mathrm{q}}{3}+\frac{\mathrm{q}}{3}+\frac{\mathrm{q} / 2}{5}+\frac{\mathrm{q} / 2}{5}\right) \\ & =\mathrm{kq}\left(\frac{10+10+3+3}{30}\right) \\ & =\mathrm{kq}\left(\frac{26}{30}\right) \\ \mathrm{V} _{\mathrm{F}} & =\mathrm{k}\left(\frac{\mathrm{q}}{2 \times 3}+\frac{\mathrm{q} / 2}{3}+\frac{\mathrm{q}}{5}+\frac{\mathrm{q}}{5}\right) \\ & =\mathrm{kq}\left(\frac{1}{6}+\frac{1}{6}+\frac{1}{5}+\frac{1}{5}\right) \\ & =\mathrm{kq}\left(\frac{5+5+6+6}{30}\right) \\ & =\mathrm{kq}\left(\frac{22}{30}\right) \end{aligned} $

$ \begin{aligned} \therefore \mathrm{V} _{\mathrm{E}} -\mathrm{V} _{\mathrm{F}} & =\mathrm{kq}\left(\frac{26-22}{30}\right)=\mathrm{kq} \times \frac{4}{30} \\ & =9 \times 10^{9} \times 10 \times 10^{-9} \times \frac{4}{30} \\ & =12 \mathrm{~V} \end{aligned} $

Correct answer is (2)

Difficult

Electric Potential

65. A long straight conducting wire is uniformly charged with density of $+10 \mathrm{mC}$ per metre. A point charge of $+2 \mu \mathrm{C}$ is released at a perpendicular distance of $10 \mathrm{~cm}$ from this wire.

Kinetic energy of this charge, when it has moved through a distance of $10 \mathrm{~cm}$, under the field of the charged wire is:

(1) $249.57 \mathrm{~J}$

(2) $108.37 \mathrm{~J}$

(3) $360.03 \mathrm{~J}$

(4) $0.25 \mathrm{~J}$

Show Answer

Correct Answer: (1)

Solution:

The kinetic energy of the particle at $4=20 \mathrm{~cm}$, when released at $\mathrm{r}=10 \mathrm{~cm}$ is:

$ \text { K.E. }=\mathrm{qV} $

Where $\mathrm{V}$ is the decrease in potential between these two points.

$~$

Now, $\mathrm{V}=\int _{\mathrm{r}=10}^{\mathrm{r}=20} \mathbf{E} \cdot \mathbf{d r}=\int _{10}^{20} \mathrm{Edr}$

$ \begin{aligned} & =\mathrm{V}=\int _{10}^{20} \frac{2 \mathrm{k} \lambda}{\mathrm{r}} \mathrm{dr}=2 \mathrm{k} \lambda\left[\log _{\mathrm{e}} \mathrm{r}\right] _{10}^{20} \quad \mathrm{E}=\frac{2 \mathrm{k} \lambda}{\mathrm{r}} \\ & =2 \mathrm{k} \lambda \times 2.303 \log _{10}\left(\frac{20}{10}\right) \\ & =2 \times 9 \times 10^{9} \times 10 \times 10^{-3} \times 2.303 \times 0.3010 \\ & =12.477654 \times 10^{7} \text { volt } \end{aligned} $

$\therefore \mathrm{KE}=\mathrm{qV}$

$ \begin{aligned} & =2 \times 10^{-6} \times 12.47765 \times 10^{7} \\ & =249.55 \mathrm{~J} \end{aligned} $

Correct answer: (1)

Difficult

Principle of Superposition of Charges

66. Ten positively charged paritcles are kept fixed on the $x-a x i s$ at points $x=10 \mathrm{~cm}, 20 \mathrm{~cm}, 30$ $\mathrm{cm}$ and son on up to $x=100 \mathrm{~cm}$. The first particle has a charge of $1.0 \times 10^{-8} \mathrm{C}$, the second $8 \times 10^{-8} \mathrm{C}$, third $27 \times 10^{-8} \mathrm{C}$ and so $\mathrm{on}$. The tenth particle has a charge of $1000 \times 10^{-8} \mathrm{C}$. The force on $1 \mathrm{C}$ of charge placed at the origin due to these ten charged particles is:

(1) $4950 \mathrm{~N}$

(2) 495 Kilo newton

(3) $4.95 \times 10^{7} \mathrm{~N}$

(4) $495 \mathrm{~N}$

Show Answer

Correct Answer: (2)

Solution:

$$ \begin{aligned} \mathrm{F} _{0} & =\mathrm{F} _{01}+\mathrm{F} _{02}+\mathrm{F} _{03}+\ldots \mathrm{F} _{010}\\ & =\frac{\mathrm{q} _{0}}{4 \pi \varepsilon _{0}}\left[\frac{\mathrm{q} _{1}}{\mathrm{r} _{1}^{2}}+\frac{\mathrm{q} _{2}}{\mathrm{r} _{2}^{2}}+\frac{\mathrm{q} _{3}}{\mathrm{r} _{3}^{2}}+\ldots . . \frac{\mathrm{q} _{10}}{\mathrm{r} _{10}^{2}}\right] \\ & =9 \times 10^{9} \times 10^{-6}\left[\frac{1 \times 10^{-8}}{(0.1)^{2}}+\frac{8 \times 10^{-8}}{(0.2)^{2}}+\frac{27 \times 10^{-8}}{(0.3)^{2}}+\ldots . . \frac{1000 \times 10^{-8}}{(1)^{2}}\right] \\ & =9 \times 10^{9} \times 10^{-6}[1+2+3+4+\ldots .10] \\ & =9 \times 10^{3} \times[55] \\ & =495 \times 10^{3} \mathrm{~N} \\ & =495 \text { Kilonewton } \end{aligned} $$

Correct answer: (2)

Easy

Electric Field

67. A charge of $Q$ is kept at common centre $O$ of conducting shells $x$ and $y$ in the figure. Charge of $2 Q$ is given to $x$ and $3 Q$ is given to $y$. The net charge enclosed by the imaginary spherical surface $z$ (represented by dotted line) is:

$~$

(1) Q

(2) $3 \mathrm{Q}$

(3) $6 \mathrm{Q}$

(4) zero

Show Answer

Correct Answer: (4)

Solution:

As the spherical surface $z$ is inside the material of the conducting shell $y, E=0$ at all points on it.

Hence flux crossing $\mathrm{z}$ is zero.

Therefore totalcharge enclosed by it also in zero.

Correct answer: (4)

Easy

Capacitor

68. A network of six identical capacitors, each of value $C$, is made as given in the diagram.

$~$

The equivalent capacitance between $A$ and $B$ is:

(1) $\frac{\mathrm{C}}{4}$

(2) $\frac{3 \mathrm{C}}{4}$

(3) $\frac{3 \mathrm{C}}{2}$

(4) $\frac{4 \mathrm{C}}{2}$

Show Answer

Correct Answer: (4)

Solution:

The given network can be re-drawn as below.

$~$

or

$~$

Or

$~$

Correct answer: (4)

Average

Capacitor

69. In the figure below, the charge on the $4 \mu \mathrm{F}$ capacitor is $24 \mu \mathrm{C}$.

$~$

The potential difference $(\mathrm{V})$ across the source is:

(1) 10 Volt

(2) 20 Volt

(3) 23.3 Volt

(4) 46.6 Volt

Show Answer

Correct Answer: (2)

Solution:

$\mathrm{C} _{1}=4 \mu \mathrm{F}$ and $\mathrm{Q} _{1}=24 \mu \mathrm{C}$

$\mathrm{V} _{1}=\frac{\mathrm{Q} _{1}}{\mathrm{C} _{1}}=\frac{24}{4}=6$ Volt

$\mathrm{C} _{2} || \mathrm{C} _{3}=1+5=6 \mu \mathrm{F}$

$~$

Since $6 \mu \mathrm{F}$ is in series with $\mathrm{C} _{1}$, charge on $6 \mu \mathrm{F}(1 || 5)$ is $24 \mu \mathrm{C}$.

$\therefore$ Potential difference across $6 \mu \mathrm{F}=\frac{24}{6}=4$ Volt.

$\mathrm{C} _{1} \sim\left(\mathrm{C} _{2} || \mathrm{C} _{3}\right)=\frac{4 \times 6}{4+6}=2.4 \mu \mathrm{F}$

Charge on this $2.4 \mu \mathrm{F}$ is $24 \mu \mathrm{C}$.

$\therefore$ Potential difference across this combination is $\frac{24 \mu \mathrm{C}}{2.4 \mu \mathrm{F}}=10 \mathrm{~V}$

Therefore potential difference across the $3.6 \mu \mathrm{F}$ capacitor is also $10 \mathrm{~V}$.

Now, $2.4 \mu \mathrm{F}$ and $3.6 \mu \mathrm{F}$ are in parallel.

$\therefore\left[\mathrm{C} _{1} \sim \mathrm{C} _{2} || \mathrm{C} _{3}\right] | \mathrm{C} _{4}=2.4+3.6=6 \mu \mathrm{F}$

With a potential difference of $10 \mathrm{~V}$.

Now $\mathrm{C} _{5}=6 \mu \mathrm{F}$ is in series with the above combination of $6 \mu \mathrm{F}$. Hence the potential difference across $\mathrm{C} _{5}$ is also $10 \mathrm{~V}$.

$\therefore$ Source supplies a potential difference of

$$ \begin{aligned} & \mathrm{V}=10 \mathrm{~V}+10 \mathrm{~V} \\ & =20 \mathrm{~V} \end{aligned} $$

Correct answer: (2)

Average

Capacitor

70. Four identical capacitors are connected together as shown in the figure. A battery of $6 \mathrm{~V}$ is connected across $A$ and $B$. If the energy stored in the combination is found to be $4.5 \mu \mathrm{J}$, each capacitor is of capacitance.

$~$

(1) $0.1 \mu \mathrm{F}$

(2) $0.625 \mu \mathrm{F}$

(3) $1.0 \mu \mathrm{F}$

(4) $5.0 \mu \mathrm{F}$

Show Answer

Correct Answer: (1)

Solution:

$\mathrm{U}=\frac{1}{2} \mathrm{C} _{\text {eff }} \mathrm{V}^{2}$

$\therefore \mathrm{C} _{\text {eff }}=\frac{2 \mathrm{U}}{\mathrm{V}^{2}}=\frac{2 \times 4.5 \times 10^{-6}}{6 \times 6}=\frac{9}{36} \times 10^{-6} \mathrm{~F}$ $=\frac{9}{36} \mu \mathrm{F}$

Now, the above combination can be re-drawn as shown.

$\therefore \mathrm{C} _{\text {eff }}=\mathrm{C}+\frac{\mathrm{C}}{2}+\mathrm{C}$

$ =\frac{5}{2} \mathrm{C} $

$~$

But $\mathrm{C} _{\text {eff }}=\frac{9}{36} \mu \mathrm{F}$

$\therefore \frac{9}{36}=\frac{5}{2} \mathrm{C}$

or $\mathrm{C}=\frac{18}{36 \times 5}=\frac{1}{10}=0.1 \mathrm{C}$

Correct answer: (1)

Easy

Capacitor

71. A $10 \mu \mathrm{F}$ capacitor is charged to a potential difference of $1000 \mathrm{~V}$. The terminals of this capacitor are disconnected from the power supply and then connected to the termainals of an uncharged $6 \mu \mathrm{F}$ capacitor. The final pontential difference across each capacitor and the the energy lost by sharing of charge, between then, are respectively.

(1) $266.6 \mathrm{~V}$ and $2.637 \mathrm{~J}$

(2) $625 \mathrm{~V}$ and $1.875 \mathrm{~J}$

(3) $250 \mathrm{~V}$ and $2.637 \mathrm{~J}$

(4) $750 \mathrm{~V}$ and $1.875 \mathrm{~J}$

Show Answer

Correct Answer: (2)

Solution:

Common potential $=\frac{\mathrm{C} _{1} \mathrm{~V} _{1}+\mathrm{C} _{2} \mathrm{~V} _{2}}{\mathrm{C} _{1}+\mathrm{C} _{2}}$

$ \begin{aligned} & =\frac{10 \mu \mathrm{F} \times 1000 \mathrm{~V}+0}{(10 \mu \mathrm{F}+6 \mu \mathrm{F})} \\ & =625 \mathrm{~V} \end{aligned} $

Energy loss $=\frac{1}{2} \frac{C _{1} C _{2}}{C _{1}+C _{2}}\left(V _{1}-V _{2}\right)^{2}$

$ \begin{aligned} & =\frac{1}{2} \times \frac{10 \times 6}{10+6}(1000-0)^{2} \mu \mathrm{J} \\ & =\frac{15}{8} \times 10^{6} \times 10^{-6} \mathrm{~J} \\ & =1.875 \mathrm{~J} \end{aligned} $

Correct answer: (2)

Average

Capacitor

72. All the capacitors in the figure have the same capacitance of $C$ each. The cell has emf $V$.

$~$

The total charge stored in the combination is:

(1) $\frac{1}{4} \mathrm{CV}$

(2) $\frac{3}{7} \mathrm{CV}$

(3) $\frac{1}{2} \mathrm{CV}$

(4) $2 \mathrm{CV}$

Show Answer

Correct Answer: (3)

Solution:

The combination can be re-drawn as shown. Since the three capacitors in the middle have no potential difference across, than, $q=0$ for each of then.

The other two are is series with $\mathrm{V}$.

$~$

$\therefore \mathrm{C} _{\text {eff }}=\frac{\mathrm{C}}{2}$

and $ ~~ \mathrm{Q}=\mathrm{C} _{\text {eff }} \times \mathrm{V}$

$ \quad =\frac{1}{2} \mathrm{CV}$

Correct answer: (3)

Average

Capacitor

73. Keeping other relevant parameters constant, which of the follwing plots is / are incorrectly drawn, in case of a capacitor. (Notation have usual meaning).

$~$

(1) (c) and (e)

(2) (b), (c) and (d)

(3) (d) and (e)

(4) (b) and (e)

Show Answer

Correct Answer: (1)

Solution:

(a) $ \quad \mathrm{U}=\frac{1}{2} \mathrm{CV}^{2}$

$\therefore \mathrm{U} \propto \mathrm{V}^{2}$

$~$

Graph of U vs V is parabola (Correct) .

(b) $ \quad \mathrm{C}=\frac{\mathrm{A} \varepsilon _{0}}{\mathrm{~d}}$

$\therefore \mathrm{C} \propto \frac{1}{\mathrm{~d}}$

$~$

Graph is correct

(c) $ \quad \mathrm{C}=\mathrm{KC} _{0}$, where $\mathrm{K}$ cannot be less than 1 .

So, $\mathrm{C} \propto \mathrm{K}$, but intercept should be $\mathrm{C} _{0}$.

$~$

Graph is incorrect.

(d) $ \quad \mathrm{U}=\overline{\mathrm{U}} \times(\mathrm{Ad})$

$\mathrm{U}=\frac{1}{2} \varepsilon _{0} \mathrm{E}^{2}(\mathrm{Ad})$

$\therefore \mathrm{U} \propto \mathrm{E}^{2}$

$~$

Graph is correct.

(e) $ \quad \mathrm{F}=\frac{1}{2} \frac{\mathrm{Q}^{2}}{\mathrm{~A}}$

$\therefore \mathrm{F} \propto \mathrm{Q}^{2}$

$~$

So graph has parabolic bending towards $\mathrm{F}$-axis is incorrect.

Correct answer: (1)

Average

Dielectric

74. $\mathbf{C}, \mathrm{V}, \mathrm{Q}, \mathrm{U}$ and $\mathrm{E}$ are respectively the capacitance, potential difference, charge, energy stored and inside electric field intensity of a parallel plate capacitor, which is connected to a battery. Now a material of dielectric constant 5 is filled between the plates of the capacitor, without disconnecting the battery. The new values of the above quantities are:

(1) $5 \mathrm{C}, 5 \mathrm{~V}, 5 \mathrm{Q}, 5 \mathrm{U}$ and $5 \mathrm{E}$

(2) 5C, V, 5Q, 5U and E

(3) $5 \mathrm{C}, \mathrm{V}, 5 \mathrm{Q}, \mathrm{U}$ and $\mathrm{E} / 5$

(4) $5 \mathrm{C}, \mathrm{V}, 5 \mathrm{Q}, 5 \mathrm{U}$ and $\mathrm{E} / 5$

Show Answer

Correct Answer: (2)

Solution:

Both the cases are represented in the diagram.

(i) Capacitance increases to $\mathrm{KC}=5 \mathrm{C}$

(ii) $\mathrm{V}$ remains same as the potential difference of the battery. $\mathrm{V}=\mathrm{V}$

(iii) Charge $=\mathrm{CV}=\mathrm{Q}$

New charge is $5 \mathrm{CV}=5 \mathrm{Q}$

$~$

(iv) $\mathrm{U} _{\mathrm{i}}=\frac{1}{2} \mathrm{CV}^{2}=\mathrm{U}$

New $\mathrm{U} _{\mathrm{f}}=\frac{1}{2} 5 \mathrm{CV}^{2}=5 \mathrm{U}$

$~$

(v) If $\mathrm{E}$ is the field inside, than

$\mathrm{V}=\mathrm{Ed}$ or $\mathrm{E}=\mathrm{V} / \mathrm{d}$

Now, $\mathrm{V}$ and d reamin the same.

$\therefore \mathrm{E}$ is same in both cases.

Correct answer: (2)

Easy

Electrostatic Force

75. Two positive ions, each carrying a charge ’ $q$ ’ are separaated by a distance ’ $d$ ‘. If ’ $F$ ’ is the force of repulsion between the ions and ’ $e$ ’ is the magnitude of electronic charge, the number of electrons missing from each ion is:

(1) $\frac{4 \pi \varepsilon _{0} \mathrm{~F}}{\mathrm{e}^{2}}$

(2) $\frac{4 \pi \varepsilon _{0} \mathrm{Fd}^{2}}{\mathrm{e}^{2}}$

(3) $\left(\frac{4 \pi \varepsilon _{0} \mathrm{Fe}^{2}}{\mathrm{~d}^{2}}\right)^{1 / 2}$

(4) $\left(\frac{4 \pi \varepsilon _{0} \mathrm{Fd}^{2}}{\mathrm{e}^{2}}\right)^{1 / 2}$

Show Answer

Correct Answer: (4)

Solution:

$\mathrm{q} _{1}=\mathrm{q} _{2}=\mathrm{q}=+\mathrm{ne}$

$\mathrm{F}=\frac{1}{4 \pi \varepsilon _{0}} \cdot \frac{\mathrm{q} _{1} \mathrm{q} _{2}}{\mathrm{~d}^{2}}$

$\mathrm{F}=\frac{1}{4 \pi \varepsilon _{0}} \cdot \frac{(\text { ne })(\text { ne })}{\mathrm{d}^{2}}$

$\therefore \mathrm{n}^{2}=\frac{4 \pi \varepsilon _{0} \mathrm{Fd}^{2}}{\mathrm{e}^{2}}$

or $\mathrm{n}=\sqrt{\frac{4 \pi \varepsilon _{0} \mathrm{Fd}^{2}}{\mathrm{e}^{2}}}$

Correct answer: (4)

Easy

Electric Dipole

76. The dipole moment, of a deuteron and an electron, separated by a distance of 1.6 femtometre is:

(1) $4.096 \times 10^{-34} \mathrm{~cm}$

(2) $5.12 \times 10^{-34} \mathrm{~cm}$

(3) $8.192 \times 10^{-34} \mathrm{~cm}$

(4) $2.56 \times 10^{-34} \mathrm{~cm}$

Show Answer

Correct Answer: (4)

Solution:

$P=q(2 a)$

$~$

Here $\mathrm{q}=\mathrm{e}=1.6 \times 10^{-19} \mathrm{C}$

$2 \mathrm{a}=1.6$ femtometre

$ =1.6 \times 10^{-15} \mathrm{~m} $

$\therefore \mathrm{P}=1.6 \times 10^{-19} \times 1.6 \times 10^{-15} \mathrm{Cm}$

$=2.56 \times 10^{-34} \mathrm{Cm}$

Correct answer: (4)

Average

Capacitors

77. A series combination of $n _{1}$ capacitors, each of value $C _{1}$, is charged by a source of potential difference $4 \mathrm{~V}$. In another case, a parallel combination of $n _{2}$ capacitors, each of value $C _{2}$, is charged by a source of potential difference $V$. If both the combinations have same total energy stored, then the value of $C _{2}$ in terms of $C _{1}$ is:

(1) $\frac{16 C _{1}}{n _{1} n _{2}}$

(2) $\frac{2 \mathrm{C} _{1}}{\mathrm{n} _{1} \mathrm{n} _{2}}$

(3) $\frac{16 C _{1} n _{2}}{n _{1}}$

(4) $\frac{2 \mathrm{C} _{1} \mathrm{n} _{2}}{\mathrm{n} _{1}}$

Show Answer

Correct Answer: (1)

Solution:

For series, $\mathrm{C} _{\mathrm{S}}=\frac{\mathrm{C} _{1}}{\mathrm{n} _{1}}$

and $\mathrm{V}=4 \mathrm{~V}$

$\therefore \mathrm{U} _{\mathrm{S}}=\frac{1}{2} \mathrm{C} _{\mathrm{S}} \mathrm{V}^{2}=\frac{1}{2} \frac{\mathrm{C} _{1}}{\mathrm{n} _{1}}(4 \mathrm{~V})^{2}$

For parallel, $\mathrm{C} _{\mathrm{P}}=\mathrm{n} _{2} \mathrm{C} _{2}$

and $\mathrm{V}=\mathrm{V}$

$\therefore \mathrm{U} _{\mathrm{S}}=\frac{1}{2} \mathrm{C} _{\mathrm{S}} \mathrm{V}=\frac{1}{2} \mathrm{n} _{2} \mathrm{C} _{2} \mathrm{~V}^{2}$

But, $\mathrm{U} _{\mathrm{S}}=\mathrm{U} _{\mathrm{P}}$

$\therefore \frac{1}{2} \frac{\mathrm{C} _{1}}{\mathrm{n} _{1}}(4 \mathrm{~V})^{2}=\frac{1}{2} \mathrm{n} _{2} \mathrm{C} _{2} \mathrm{~V}^{2}$

or $\quad \mathrm{C} _{2}=\frac{16 \mathrm{C} _{1}}{\mathrm{n} _{1} \mathrm{n} _{2}}$

Correct answer: (1)

Average

Electrostatic Potential Energy

78. Three identical charges, each of $2 \mu \mathrm{C}$, are placed at the vertices of a triangle $\mathrm{ABC}$. If $\mathrm{AB}+\mathrm{BC}=12 \mathrm{~cm}$ and $\mathrm{AB} \cdot \mathrm{BC}=32 \mathrm{~cm}^{2}$, the potential energy of the charge at $B$ is:

(1) $1.53 \mathrm{~J}$

(2) $5.31 \mathrm{~J}$

(3) $3.15 \mathrm{~J}$

(4) $1.35 \mathrm{~J}$

Show Answer

Correct Answer: (4)

Solution:

Given,

$\mathrm{AB}+\mathrm{BC}=12 \mathrm{~cm}$ and $\mathrm{AB} \cdot \mathrm{BC}=32 \mathrm{~cm}^{2}$

$~$

Now, $(A B-B C)^{2}=(A B+B C)^{2}-4 A B \cdot B C$

$ \begin{aligned} & =144-128 \\ & =16 \mathrm{~cm}^{2} \end{aligned} $

$\therefore \mathrm{AB}-\mathrm{BC}=4 \mathrm{~cm}$

Hence $\mathrm{AB}=8 \mathrm{~cm}, \mathrm{BC}=4 \mathrm{~cm}$

The potential $V$ at $B$ is:

$\mathrm{V}=\mathrm{Kq}\left(\frac{1}{8 \times 10^{-2}}+\frac{1}{4 \times 10^{-2}}\right)$

$ \begin{aligned} & =\frac{9 \times 10^{9} \times 2 \times 10^{-16}}{8 \times 10^{-2}} \times 3 \\ & =\frac{27}{4} \times 10^{+5} \mathrm{Volt} \end{aligned} $

Now, $\mathrm{U}=\mathrm{qV}$

$ \begin{aligned} & =2 \times 10^{-6} \times \frac{27}{4} \times 10^{+5} \mathrm{~J} \\ & =\frac{27}{2} \times 10^{-1} \mathrm{~J}=1.35 \mathrm{~J} \end{aligned} $

Correct answer: (4)

Easy

Electrostatic Charge

79. Two bodies A and B carry charges of $-3.00 \mu \mathrm{C}$ and $-0.44 \mu \mathrm{C}$ respectively. How many electrons should be transferred, from which one to which, so that they acquire equal charges?

(1) $8 \times 10^{18}$, from A to $\mathrm{B}$

(2) $8 \times 10^{12}$, from $A$ to $B$

(3) $8 \times 10^{13}$, from $\mathrm{A}$ to $\mathrm{B}$

(4) $4 \times 10^{12}$, from $A$ to $B$

Show Answer

Correct Answer: (2)

Solution:

A has higher negative charge and so electrons should be transferred from A to B.

Let $\mathrm{n}$ be the number of electrons transferred to make the charges equal.

$\therefore-3.00+\mathrm{ne}=-0.44-\mathrm{ne}\qquad \quad \qquad $ (in $\mu \mathrm{C}$, where $\mathrm{e}=1.6 \times 10^{-13} \mu \mathrm{C}$ )

or 2 ne $=2.56 \qquad \quad \qquad \qquad\qquad \qquad $ (in $\mu \mathrm{C}$ )

$\therefore \mathrm{n}=\frac{2.56}{2 \times \mathrm{e}} \mu \mathrm{C}$

$=\frac{2.56}{2 \times 1.6 \times 10^{-13}} \mu \mathrm{C}$

$ \begin{aligned} & =0.8 \times 10^{13} \\ & =8 \times 10^{12} \end{aligned} $

Correct answer: (2)

Average

Electrostaic Force

80. Two identical helium filled balloons $A$ and $B$, charged with ’ $q$ ’ each and fastened to a body of $60 \mathrm{~g}$, by threads of length $5 \mathrm{~cm}$ each. Float in air in equilibrium, as shown in the figure. If the seperation between the two balloons (assumed spherical) is $6 \mathrm{~m}$, the charge $q$ is: (Take $g$ $=\mathbf{1 0} \mathbf{m s}^{-2}$ )

$~$

(1) $30 \mu \mathrm{C}$

(2) $\sqrt{1.5} \times 10^{-5} \mathrm{C}$

(3) $15 \mu \mathrm{C}$

(4) $45 \mu \mathrm{C}$

Show Answer

Correct Answer: (1)

Solution:

At equilibrium

$\mathrm{T} \sin \theta=\mathrm{Fe}, \mathrm{T} \cos \theta=\frac{\mathrm{mg}}{2}$

$\therefore \tan \theta=\frac{2 \mathrm{Fe}}{\mathrm{mg}}$

$~$

$\tan \theta=\frac{3}{4} ; \mathrm{Fe}=\frac{\mathrm{kq}^{2}}{6^{2}} ; \mathrm{mg}=60 \times 10^{-3} \times 10$

$\therefore \frac{3}{4}=2 \times \frac{\mathrm{kq}^{2}}{6^{2}} \times \frac{1}{60 \times 10^{-3} \times 10}=\frac{2 \times 9 \times 10^{9} \times \mathrm{q}^{2}}{6^{2} \times 6 \times 10^{-1}}$

$\therefore \mathrm{q}^{2}=0.9 \times 10^{-9}=9 \times 10^{-10} \mathrm{C}^{2}$

$\mathrm{q}=3 \times 10^{-5} \mathrm{C}=30 \mu \mathrm{C}$

Correct answer: (1)

Average

Electric Dipole

81. A dipole of very small length and dipole moment $P=5 \times 10^{-7} \mathrm{~cm}$, lies at the origin along the positive $x$-axis. A point $A$ is located at $(3,4) \mathrm{m}$. The magnitude of the electric field, of the dipole, at point $A$ is:

(1) $36 \mathrm{NC}^{-1}$

(2) $36 \sqrt{13} \mathrm{NC}^{-1}$

(3) $14.4 \sqrt{13} \mathrm{NC}^{-1}$

(4) $14.4 \sqrt{52} \mathrm{NC}^{-1}$

Show Answer

Correct Answer: (3)

Solution:

$$ \begin{aligned} E=\frac{k P}{r^{3}} & \sqrt{1+3 \cos ^{2} \theta} \\ & =\frac{9 \times 10 \times 5 \times 10^{-7}}{(5)^{3}} \sqrt{1+3 \times\left(\frac{3}{5}\right)^{2}} \\ & =7.2 \times 5 \sqrt{\frac{25+27}{25}} \\ & =7.2 \times 2 \sqrt{13} \\ & =14.4 \sqrt{13} \mathrm{NC}^{-1} \end{aligned} $$

$~$

Difficult

Electrostatic Potential Energy

82. The electrostatic energy of a nucleous can be considered as the work done against electrostatic forces in assemblying the $z$ protons in a sphere of radius $R$.

Therefore the electrostatic energy assocaited with a nucleus is given by:

(1) $\frac{\mathrm{z}^{2} \mathrm{e}^{2}}{4 \pi \varepsilon _{0} \mathrm{R}}$

(2) $\frac{\mathrm{ze}}{4 \pi \varepsilon _{0} \mathrm{R}^{2}}$

(3) $\frac{3}{5} \frac{\mathrm{z}^{2} \mathrm{e}^{2}}{4 \pi \varepsilon _{0} \mathrm{R}}$

(4) $\frac{\mathrm{z}^{2} \mathrm{e}^{2}}{8 \pi \varepsilon _{0} \mathrm{R}}$

Show Answer

Correct Answer: (2)

Solution:

The uniform density of charge in a nucelus is given by:

$$ \rho=\frac{\mathrm{ze}}{\frac{4}{3} \pi \mathrm{R}^{3}} $$

We regard the nucleus as a sphere having a uniform charge distribution.

The work done to increase in radius of sphere $r$ to a sphere of radius $r+d r$ is given by work done in bringing charge dq present in a spherical shell of thickness dr from infinity to $r$ against the charge on sphere of radius $r$.

$\mathrm{dq}=3 \frac{\mathrm{ze}}{\mathrm{R}^{3}} \mathrm{r}^{2} \mathrm{dr} \times \frac{\frac{4}{3} \pi \mathrm{r}^{2} \rho}{4 \pi \varepsilon _{0} \mathrm{r}}$

$\therefore \mathrm{dq}=\rho \times 4 \pi \mathrm{r}^{2} \mathrm{dr}$

$\left(\right.$ or $\left.\mathrm{dq}=4 \pi \mathrm{r}^{2} \mathrm{dr}=\left(3 \frac{\mathrm{ze}}{\mathrm{R}^{3}}\right) \mathrm{r}^{2} \mathrm{dr}\right)$

Putting $\rho=\frac{\text { ze }}{\frac{4}{3} \pi R^{3}}$

$\mathrm{dv}=\frac{3 \mathrm{z}^{2} \mathrm{e}^{2}}{4 \pi \varepsilon _{0} \mathrm{R}^{6}} \pi \mathrm{r}^{4} \mathrm{dr}$

$\therefore \mathrm{U}=\int _{0}^{\mathrm{R}} \frac{3 \mathrm{z}^{2} \mathrm{e}^{2}}{4 \pi \varepsilon _{0} \mathrm{R}^{6}} \mathrm{r}^{2} \mathrm{dr}$

or $\quad \mathrm{U}=\frac{3}{5} \frac{\mathrm{z}^{2} \mathrm{e}^{2}}{4 \pi \varepsilon _{0} \mathrm{R}}$

Correct answer: (2)

Difficult

Electrostatic Potential Energy

83. A conducting sphere of radius $R$ is having a surface density $\sigma$. What is the energy stored in the surrounding space due to this charge distribution on conducting sphere? (The charge on the sphere is denoted by $q$ )

(1) $\frac{\mathrm{q}^{2}}{8 \pi \varepsilon _{0} \mathrm{R}}$

(2) $\frac{3}{5} \frac{q^{2}}{8 \pi \varepsilon _{0} R}$

(3) $\frac{\mathrm{q}^{2}}{4 \pi \varepsilon _{0} \mathrm{R}}$

(4) $\frac{q^{2}}{8 \pi \varepsilon _{0} R^{2}}$

Show Answer

Correct Answer: (1)

Solution:

Let us consider a point $P$ at distance $r$ from the centre of the sphere $(r>R)$

The clectric field at $\mathrm{P}$ is give by

$\mathrm{E}=\frac{1}{4 \pi \varepsilon _{0}} \frac{\mathrm{q}}{\mathrm{r}^{2}}$

Let the energy density at this point $\mathrm{P}$ is given by

$\mathrm{u}=\frac{1}{2} \varepsilon _{0} \mathrm{E}^{2}=\frac{1}{2} \varepsilon _{0} \frac{\mathrm{q}^{2}}{16 \pi^{2} \varepsilon _{0}^{2}}$

The energy lying in a spherical slab formed between two concentric spheres of radii $\mathrm{r}$ and $\mathrm{r}+\mathrm{dr}$ is $\mu \lambda 4 \pi \mathrm{r}^{2} \mathrm{dr}$ denote this energy by $\mathrm{dU}$ we take

$$ \begin{aligned} & d U=4 \pi r^{2} d r \frac{q^{2}}{32 \pi^{2} \varepsilon _{0} r^{4}} \\ & U=\int _{r=R}^{r=\alpha} d u=\frac{q^{2}}{8 \pi \varepsilon _{0}} \int _{R}^{\infty} \frac{d r}{r^{2}} \\ &=\frac{q^{2}}{8 \pi \varepsilon _{0} R} \end{aligned} $$

Correct answer: (1)

Easy

Dielectric

84. A polar dielectric is placed in a external field $E _{e}$. Which of the following statement is correct?

(1) The external field $\mathbf{E} _{\mathrm{e}}$ will not effect the dielectric in any way what so ever.

(2) The external field will allign all the atoms and molecules of the given dielectric in its own direction.

(3) The dielectric will be have as if it is equivlent to a large number of tiny dipoles situated in vaccum.

(4) The dielectric will behave as if it is equivalent to a large number of tiny dipoles situated in the material of dielectric.

Show Answer

Correct Answer: (3)

Solution:

A polar dielectric has a built in dipole moment. In the absence of an external field these dipoles are orineted complete by at randomly. The net dipole moment associated with any macroscoping small volume element is zero. When an electric field is applied there is is a small induced dipole moment and also a torque is exerted on all permanent dipole and the field tends to allign the dipoles in its own direction. Thermal motion tends to oppose this alignment so that at given temperature some alignment in the direction of the field is produced in equilibrium condition. Due to this aligment there is now some net dipole moment associated with each macroscoping small element of the dielectric. Hence a dielectric placed in an external field can be considered as equivalent to a large number of tiny dipole stituated in vacuum.

Correct answer: (3)

Easy

Dielectric

85. In the reaction $\mathbf{P}=\varepsilon _{0} x _{\mathrm{e}} \mathbf{E}$

For a linear dielectric, while of the following statement is correct $P$.

(1) $\mathbf{E}$ stand for external field of any value in which the dielectric is placed.

(2) $\mathbf{E}$ stand for external field of very large values in which the dielectric is placed.

(3) $\mathbf{E}$ stands for the field produced by dipoles to which a give dielectric is equivelent to.

(4) $\mathbf{E}$ stands for the vector sum of the external field and the secondary field produced by dipoles to which a given dielectric is equivalent to.

Show Answer

Correct Answer: (4)

Solution:

The statement (4) is established by experiment.

Average

Dielectric

86. If $\varepsilon _{0}$ stand for the permitivity of vacuum, and $\varepsilon$ the permitivity of a given dielectric, which of the follows relation is correct?

(1) $\mathbf{P}=\leftarrow \mathbf{E}$

(2) $\mathbf{P}=\left(\varepsilon-\varepsilon _{0}\right) \mathbf{E}$

(3) $\mathbf{P}=\left(\frac{\varepsilon-\varepsilon _{0}}{\varepsilon _{0}}\right) \mathbf{E}$

(4) $\mathbf{P}=\left(\frac{\varepsilon}{\varepsilon _{0}}\right) \mathbf{E}$

Show Answer

Correct Answer: (2)

Solution:

We have,

$\mathbf{D}=\varepsilon _{0} \mathbf{E}+\mathbf{P}$

or $\quad \mathbf{D}=\varepsilon _{0} \mathbf{E}+\varepsilon _{0} x _{\mathrm{e}} \mathbf{E}$

$ =\varepsilon _{0} \mathbf{E}\left(1+x _{\mathrm{e}}\right) $

Now, $\varepsilon=\varepsilon _{0}\left(1+x _{\mathrm{e}}\right)$

$$ \begin{align*} & \mathbf{D}=\varepsilon \mathbf{E} \qquad\qquad \ldots . \ldots {(1)}\\ & \mathbf{P}=\varepsilon _{0} x _{\mathrm{e}} \mathbf{E} \qquad\qquad \ldots . \ldots {(2)}\\ & \varepsilon=\varepsilon _{0}\left(1+x _{\mathrm{e}}\right) \qquad \qquad \ldots . \ldots {(3)} \end{align*} $$

From equations 1, 2 and 3 we obtain.

$$ \mathbf{P}=\left(\varepsilon-\varepsilon _{0}\right) \mathbf{E} $$

Correct answer: (2)