INTRODUCTION

Electricity is the cause of something present in matters called the ’electric charge’. Although charges are present in all matters but all of them do not produce electricity. Actually, electricity is produced by those matters which are ’electrically chas ged’ or by charges in motion. The electricity produced by charges at rest is known as static electricity and the electricity produced by charges in motion is called ‘current electricity’.

This chapter deals with the electricity produced by charges at rest. A static charge can produce an electric field but it can’t produce an electric current. Electric current can be produced only when the charges are in motion which is the subject matter of the next chapter. Electric charges can be stored and used as per requirement. The device which stores charges is known as ‘capacitor’ to be dealt with in this chapter.

Electric charge

We know that mass is responsible of gravitational force. Similarly, there is a property to every massive object which is responsible for another kind of force which can be attractive or repulsive unlike gravitational force which is only attractive. Also, this force is much more stronger than gravitational forces. This force is termed as electrical force. So, charge is responsible for electrical forces.

Electric eels can generate an electrical charge of up to 600 volts in order to stun prey and keep predators at bay. Up to 6,000 electroplates are arranged like a dry cell in the eel’s body. Its internal organs are all in a small area behind the head, with 7/8 of the eel being tail. The electrical shocks come from muscles mainly in the tail portion of the electric eel’s body. The body of an electric eel is similar to a battery. The tail end of the eel has a positive charge and the head region is negatively charged. When the eel touches its tail and head to other animals it sends electric shocks through their bodies. When the eel is at rest, there is no generation of electrical impulses.

Fig. 3.1 : Electric eels

HISTORICAL PREVIEW

It was discovered centuries ago that certain types of materials would mysteriously attract one another after being rubbed together. For example: after rubbing a piece of silk against a piece of glass, the silk and glass would tend to stick together. Indeed, there was an attractive force that could be demonstrated even when the two materials were separated:

Glass and silk aren’t the only materials known to behave like this. Anyone who has ever brushed up against a latex balloon only to find that it tries to stick to them has experienced this same phenomenon. Paraffin wax and wool cloth are another pair of materials early experimenters recognized as manifesting attractive forces after being rubbed together.

This phenomenon became even more interesting when it was discovered that identical materials, after having been rubbed with their respective cloths, always repelled each other:

Glass rod

(a)

It was also noted that when a piece of glass rubbed with silk was exposed to a piece of wax rubbed with wool, the two materials would attract one another:

Furthermore, it was found that any material demonstrating properties of attraction or repulsion after being rubbed could be classed into one of two distinct categories: attracted to glass and repelled by wax, or repelled by glass and attracted to wax. It was either one or the other: there were no materials found that would be attracted to or repelled by both glass and wax, or that reacted to one without reacting to the other.

More attention was directed toward the pieces of cloth used to do the rubbing. It was discovered that after rubbing two pieces of glass with two pieces of silk cloth, not only did the glass pieces repel each other, but so did the cloths. The same phenomenon held for the pieces of wool used to rub the wax:

Now, this was really strange to witness. After all, none of these objects were visibly altered by

by the rubbing, yet they definitely behaved differently than before they were rubbed. Whatever change took place to make these materials attract or repel one another was invisible.

Some experimenters speculated that invisible “fluids” were being transferred from one object to another during the process of rubbing, and that these “fluids” were able to effect a physical force over a distance. Charles Dufay was one the early experimenters who demonstrated that there were definitely two different types of changes brought by rubbing certain pairs of objects together. The fact that there was more than one type of change manifested in these materials was evident by the fact that there were two types of forces produced: attraction and repulsion. The hypothetical fluid transfer became known as a charge.

One pioneering researcher, Benjamin Franklin, came to the conclusion that there was only one fluid exchanged between rubbed objects, and that the two different “charges” were nothing more than either an excess or a deficiency of that one fluid. After experimenting with wax and wool, Franklin suggested that the coarse wool removed some of this invisible fluid from the smooth wax, causing an excess of fluid on the wool and a deficiency of fluid on the wax. The resulting disparity in fluid content between the wool and wax would then cause an attractive force, as the fluid tried to regain its former balance between the two materials. Postulating the existence of a single “fluid” that was either gained or lost through rubbing accounted best for the observed behavior: that all these materials fell neatly into one of two categories when rubbed, and most importantly, that the two active materials rubbed against each other always fell into opposing categories as evidenced by their invariable attraction to one another. In other words, there was never a time where two materials rubbed against each other both became either positive or negative.

Following Franklin’s speculation of the wool rubbing something off of the wax, the type of charge that was associated with rubbed wax became known as “negative” (because it was supposed to have a deficiency of fluid) while the type of charge associated with the rubbing wool became known as “positive” (because it was supposed to have an excess of fluid). Little did he know that his innocent conjecture would cause much confusion for students of electricity in the future.

Precise measurements of electrical charge were carried out by the French physicist Charles Coulomb in the 1780’s using a device called a torsional balance measuring the force generated between two electrically charged objects. The results of Coulomb’s work led to the development of a unit of electrical charge named in his honor, the coulomb. If two “point” objects (hypothetical objects having no appreciable surface area) were equally charged to a measure of I coulomb, and placed I meter (approximately I yard) apart, they would generate a force of about 9 billion newtons (approximately 2 billion pounds), cither attracting or repelling depending on the types of charges involved.

It was discovered much later that this “fluid” was actually composed of extremely small bits of matter called electrons, so named in honour of the ancient Greek word for amber: another material exhibiting charged properties when rubbed with cloth Experimentation has since revealed that all objects are composed of extremely small “building-blocks” known as atoms, and that these atoms are in turn composed of smaller components known as particles. The three fundamental particles comprising atoms are called protons, neutrons, and electrons.

The result of an imbalance of this “fluid” (electrons) between objects is called static electricity. It is called “static” because the displaced electrons tend to remain stationary after being moved from one material to another.

Activity : Take any two materials from the following list and then rubbed with each other. We can always find that the former one is positively charged and the later one is negatively charged.

Fur $\to$ glass $\to$ paper $\to$ metal $\to$ silk $\to$ plastic $\to$ amber $\to$ rubber $\to$ sulfur

When a charged body is close enough to a neutral body, they attract each other. One of the applications of this effect is to use tiny paint droplets to paint the automobiles uniformly.

CHECK POINT :

When one material is rubbed against another, electrons jump readily from one to the other, but protons do not. Why is this? (Think in atomic terms.)

Show Answer

SOLUTION

Protons are bound with the nucleus of an atom but electrons revolve round it. Therefore, electrons can easily be dislodged from an atom but it is difficult to take out a proton from the atom. This is why, when one material is nubbed against another, electrons jump readily from one to the other, but protons do not.

PROPERTIES OF CHARGES :

1. Charge is conserved

That is, net charge can neither be created nor destroyed. Note the term net charge. Although charges can be created or destroyedin some processes but the net charge always remains conserved.

2. Charge is quantized

Charge on any particle can never be less than the charge of an electron (i.e., $e=-1.6 \times 10^{-19} C$ ).

In other words, the total charge on any particle is an integral multiple of the charge of electron $(e)$. Let a particle has charge $Q$ then $q= \pm n e$, where $n=1,2,3,4, \ldots \ldots$.

It means that a charge of $\frac{2}{3} e, \frac{1}{4} e$ or $\frac{4}{5} e$ on any body is not possible as they are not integral multiple of the charge of electron.

3. Unlike charges repel each other and like charges attract each other. 4. A charge at rest produces an electric field in the space around it. A uniformly moving charge produces both electric and magnetic field and a non-uniformly moving charge radiates energy.

ILLUSTRATION-3.1

Calculate the net charge on a substance consisting of (a) $5 \times 10^{14}$ electrons (b) a combination of $7 \times 10^{13}$ protons and $4 \times 10^{13}$electrons.

Show Answer

SOLUTION:

(a) The charge of one electron is $-1.6 \times 10^{-19} C$. So net charge on a substance consisting of $5 \times 10^{14}$ electrons is$5 \times 10^{14} \times(-1.6 \times 10^{-19} C)=-8 \times 10^{-5} C=-80 \mu C$.

(b) Similarly the net charge on a substance consisting of a combination of $7 \times 10^{13}$ protons and $4 \times 10^{13}$ electron is $[7 \times 10^{13} \times(1.6 \times 10^{-19} C)]+[4 \times 10^{13}(-1.6 \times 10^{-19}-C].$

$=+4.8 \mu C$.

( $\because$ the charge on one proton is $+1.6 \times 10^{19}$ )

ILLUSTRATION-3.2

When a piece of polythene is rubbed with wool, a charge of $-2 \times 10^{-7} C$ is developed on polythene. What is the amount of mass, which is transferred to polythene?

Show Answer

SOLUTION:

No. of electrons transferred, $n=\frac{q}{e}$

Mass transferred $=m_e \times n=m_e \times(\frac{\hat{q}}{e})$

$ \begin{aligned} & =9.1 \times 10^{-31} \times(\frac{2 \times 10^{-7}}{1.6 \times 10^{-19}}) \\ & =11.38 \times 10^{-19} kg \end{aligned} $

CONDUCTORS AND INSULATORS

Suppose you charge a rubber rod and then touch it to a neutral object. Some charge, repelled by the negative charge on the rod, will be transferred to the originally-neutral object. What happens to that charge then depends on the material of which the originally-neutral object consists. In the case of some materials, the charge will stay on the spot where the originally neutral object is touched by the charged rod. Such materials are referred to as insulators, materials through which charge cannot move, or, through which the movement of charge is very limited. Examples of good insulators are quartz, glass, and air. In the case of other materials, the charge, almost instantly spreads out all over the material, in response to the force of repulsion (recalling that force causes acceleration which leads to the movement) that each elementary particle of the charge exerts on every other elementary particle of charge. Materials in which the charge is free to move about are referred to as conductors. Examples of good conductors are metals and saltwater.

When you put some charge on a conductor, it immediately spreads out all over the conductor. The larger the conductor, the more it spreads out. In the case of a very large object, the charge can spread out so much that any chunk of the object has a negligible amount of charge and hence, behaves as if were neutral. Near the surface of the earth, the earth itself is large enough to play such a role. If we bury a good conductor such as a long copper rod or pipe, in the earth, and connect to it another good conductor such as a copper wire, which we might connect to another metal object, such as a cover plate for an electrical socket, above but near the surface of the earth, we can take advantage of the earth’s nature as a huge object made largely of conducting material. If we touch a charged rubber rod to the metal cover plate just mentioned, and then withdraw the rod, the charge that is transferred to the metal plate spreads out over the earth to the extent that the cover plate is neutral. We use the expression “the charge that was transferred to the cover plate has flowed into the earth.” A conductor that is connected to the earth in the manner that the cover plate just discussed is connected is called “ground.” The act of touching a charged object to ground is referred to as grounding the object. If the object itself is a conductor, grounding it (in the absence of other charged objects) causes it to become neutral.

CHARGING BY INDUCTION

When a charged particle is taken near to neutral metallic object then the electrons move to one side and there is excess of electrons on that side making it negatively charged and deficiency on the other side making that side positively charged. Hence charges appear on two sides of the body (although total charge of the body is still zero). This phenomenon is called induction and the charge produced by it is called induced charge.

A body can be charged by means of (a) friction, (b) conduction, (c) induction, (d) thermoionic ionisation, (e) photoelectric effect and (f) field emission.

COULOMB’S LAW

The force exerted by one point charge on another acts along the line joining the two charges. It varies inversely as the square of the distance separating the charges and is proportional to the product of the charges. The force depends on the medium. The force is repulsive if the charges have the same sign and attractive if the charges have opposite signs.

The magnitude of the electric force exerted by a charge $q_1$ on another charge $q_2$ a distance $r$ away is thus given by :

$ F=\frac{k|q_1 q_2|}{r^{2}} $

Where $k$ is an experimentally determined constant called the Coulomb constant, shich has the value

$ k=8.99 \times 10^{9} N \cdot m^{2} / C^{2} $

If $q_1$ is at position $ \vec{r} _1$ and $q_2$ is at $ \vec{r} _2$, the force $ \vec{F} _{1,2}$ exerted by $q_1$ on $q_2$ is

$ \vec{F} _{1,2}=\frac{k q_1 q_2}{r_1^{2}, 2} \vec{r} _{1,2} $

Where $ \vec{r} _{1,2}= \vec{r} _2- \vec{r} _1$ is the vector pointing from $q_1$ to $q_2$, and $\hat r _{1,2}= \vec{r} _{1,2} / r _{1,2}$ is a unit vector pointing from $q_1$ to $q_2$.

It is common practice to express $k$ in terms of another constant $\varepsilon_0$, by writing $k=1 /(4 \pi \varepsilon_0) ; \varepsilon_0$ is called the permittivity of free space and has a value of $\varepsilon_0=1 /(4 \pi k)=8.85 \times 10^{-12} C^{2} /(N / m^{2})$. i.e. per mission of activity between charge depends on medium.

If some dielectric (insulator) is present in the space between the charges, the net force acting on each charge is altered. The force decreases $k$ times if the medium extends till infinity. Here $k$ is a dimensionless constant which depends on the medium and called dielectric constant of the medium. Thus

$ F_e=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q_1 q_2}{r_2} \text{ or } \quad F_e^{\prime}=\frac{F_e}{k}=\frac{1}{4 \pi \varepsilon_0 k} \cdot \frac{q_1 q_2}{r_2}=\frac{1}{4 \pi \varepsilon} \cdot \frac{q_1 q_2}{r^{2}} $

Here, $\varepsilon=\varepsilon_0 k$ is called permittivity of the medium.

For vacuum or air $K=1$, For water $k=80$, For Mica $k=7$ to 10

CHECK POINT:

How does the magnitude of electric force compare with the charge between a pair of charged particles when they are brought to half their original distance of separation? To one-quarter their original distance? To four times their original distance? (What law guides your answers?)

Show Answer

SOLUTION:

According to inverse-square law, the electric force between two charges is inversely proportional to the square of the distance of their separation. Therefore, when the charges are brought to half their original distance of separation, the electric force between them increases by four times.

If the separation distance is made one-quarter their original distance, the force increases by sixteen times.

Similarly, the force between the charges decreases by sixteen times when the separation distance between them is made four times their original distance. (The inverse square law guides our answer).

ILLUSTRATION-3.3

Three identical point charges are placed at the corners of an equilateral triangle. If the electrostatic force between the any towcharges is $F$, then find the net electrostatic force on the each one.

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SOLUTION:

Three identical points charges, say, positive and having magnitude %, ure placed at the corners of’ an equilateral triangle, as shown. The change at the top conner experiences two electrostatic repulsive forces $f$, fid do the other two charges. Then, the net electrostatic force on this charge becomes

and acts in direction perpendicular to the base of triangle, as shown.

ILLUSTRATION-3.4

Tow points charges $q_1=+2 C$ and $q_2=-I C$ are separated by a distance $r$. Then, define the position of a third point charge $q_3=+I C$ in the equilibrium on the line passing through the two charges.

Show Answer

SOLUTION:

The third charge $q_3$ must be situated o:. the line passing through the two charges $q_1$ and $q_2$, as shown. This charge experiences a repulsive electrostatic force $F _{31}$ due to $q_1$ and an attractive electrostatic force $F _{32}$ due to $q_2$. The forces $F _{31}$ and $F _{32}$ are equal but opposite so that the net electrostatic fore on the charge $\eta_3$ is zero and hence, it is in the equilibrium.

$ \begin{aligned} & F _{31}=F _{32} \\ \Rightarrow & (\frac{1}{4 \pi \varepsilon_0}) \frac{2 \times 1}{(r+r^{\prime})^{2}}=(\frac{1}{4 \pi \varepsilon_0}) \frac{1 \times 1}{(r^{\prime})^{2}} \\ \Rightarrow & r+r^{\prime}=\sqrt{2} r^{\prime} \\ \Rightarrow & r^{\prime}=\frac{r}{(\sqrt{2}-1)} \end{aligned} $

which is the distance of $q_3$ from $q_2$.

ILLUSTRATION-3.5

Electric force between two point charges $q$ and $Q$ at rest is $F$. Now if a charge $-q$ is placed next to $q$ what will be the (a) force on $Q$ due to $q(b)$ total force on $Q$ ?

Show Answer

SOLUTION:

(a) As electric force between two body interaction, i.e., force between two particles, is independent of presence or absence of other particles, the force between $Q$ and $q$ will remain unchanged, i.e., $F$.

(b) An electric force is proportional to the magnitude of charges, total force on $Q$ will be given by :

$\frac{F^{\prime}}{F}=\frac{Q q^{\prime}}{Q q}=\frac{q^{\prime}}{q}=\frac{0}{q}=0 \quad[.$ as $.q^{\prime}=q+(-q)=0]$

i.e., The resultant force on $Q$ will be zero.

ELECTRIC FIELD

If we place a charge in any region of space, it is observed that any other charge placed in that region experiences a force. The space around an electric charge is filled with a field called electric field.

The intensity of electric field is a vector quantity which at a point, is defined as the force experienced by a unit positive charge placed at that point. Its direction is in the direction of the electric field if the charge is positive and opposite to it if the charge is negative. Mathematically,

$ E=\frac{F}{q}, \text{ where } E=\text{ electric field; } F=\text{ force experienced by charge } q \text{. } $

The unit of electric field is newton per coulomb (N/C).

ILLUSTRATION-3.6

A particle of mass $m$ and charge $+q$ is placed at rest in a uniform electric field $\vec{E}$, as a shown, and released. Calculate the kinetic energy it gains after moving a distance $h$.

Show Answer

SOLUTION:

The particle experiences an electric force of magnitude

in the downward direction. This force does a positive work $W=F \times h=q E h$

on the particle and the particle gains a kinetic energy equal to this work done, when it is released in the electric field and moves downwards through the distance $h$ due to the electric force on it.

Electric field due to Single Point Charge

The electric field $\vec{E}$ at a point $P$ at a distance $r$ due to a single point charge $q$ has the magnitude

$E=\left(\frac{1}{4 \pi \varepsilon_0 K}\right)\left(\frac{q}{r^2}\right) \Rightarrow E \propto \frac{1}{r^2}$

and direction along the line joining the point charge $q$ and point $P$, being away for the positive charge and towards for the negative charge.

Electric field due to Many Point Charges

The electric field due to many point charges $q_j \equiv q_1, q_2, q_3$,

at a point $P$ at the distances $r_j=r_1, r_2, r_3, r_4$,

from the charges $q_j$ is given by the superposition principle, i.e.,

$ \vec{E}= \vec{E} _1+ \vec{E} _2+ \vec{E} _3+ \vec{E} _4+\ldots \ldots \ldots \ldots=\sum_j \vec{E} _j $

where $j=1,2,3,4$, The magnitude $E_j$ due to a single point charge $q_j$ at the distance $r_j$ is

$ E_j=(\frac{1}{4 \pi \varepsilon_0 K})(\frac{q_j}{r_j^{2}}) $

For example, the electric field at the origin $(x=0)$ due to an infinite number of positive point charges, each of the magnitude $q$. situated at the points $x=1 m, 2 m, 4 m, \ldots .$. has the magnitude

$ \begin{aligned} E & =(\frac{1}{4 \pi \varepsilon_0 K})(\frac{q}{1^{2}}+\frac{q}{2^{2}}+\frac{q}{4^{2}}+\frac{q}{8^{2}}+\ldots . . .. \\ & =(\frac{q}{4 \pi \varepsilon_0 K})(1+\frac{1}{4}+\frac{1}{16}+\frac{1}{64}+\ldots \ldots \ldots . . .)=(\frac{q}{4 \pi \varepsilon_0 K})(\frac{1}{1-1 / 4})=\frac{4}{3}(\begin{matrix} q \\ 4 \pi \varepsilon_0 K \end{matrix} ) \end{aligned} $

and the minus $x$-direction.

If the infinite number of point charges are alternately positive and negative, then the electric field has the magnitude

$ \begin{aligned} E & =(\frac{1}{4 \pi \varepsilon_0 K})(\frac{q}{1^{2}}-\frac{q}{2^{2}}+\frac{q}{4^{2}}-\frac{q}{8^{2}}+\ldots \ldots .) \\ & =(\frac{q}{4 \pi \varepsilon_0 K})(1-\frac{1}{4}+\frac{1}{16}-\frac{1}{64}+\ldots \ldots) \\ & =(\frac{q}{4 \pi \varepsilon_0 K})(\frac{1}{1+1 / 4})=\frac{4}{5}(\frac{q}{4 \pi \varepsilon_0 K}) \end{aligned} $

and the minus $x$-direction.

Charged Conductor Carrying no Currents

Any charge given to a conductor spreads over the whole external-most surfaces $S$ of the conductor, with the varying surface charge density $\sigma$. The surface charge density $\sigma$ is more on the bulging parts and less on the flat parts of the external-most surface. No part of the charge given resides over the inner surfaces, $S_1, S_2 \ldots \ldots .$. etc. of the conductor, i.e., the cavity surfaces, if any.

The electric field $\vec{E}$ is vanishing inside of the external-most surface $S$ and, even, inside of the inner surfaces $S_1, S_2, \ldots$ of the conductor. The field is nonvanishing on and near the surface $S$ in the direction normal to it, being away from the conductor for the positive charge and towards for the negative charge.

$ E=\frac{\sigma}{\varepsilon_0 K} $

which shows that the field is independent of the position and depends on the surface charge density $\sigma$.

ELECTRIC LINES OF FORCE OR ELECTRIC FIELD LINES

(i) The concept of electric field was introduced by Michael Faraday.

The magnitude of electric field strength at any point is measured by the number of electric lines of force passing per unit small area around that point normally and the direction of field at any point is given by the tangent to the line of force at the point.

(ii) An electric line of force is that imaginary smooth curve drawn in an electric field along which a free isolated unit positive (initially at rest) charge moves.

Properties:

(1) The lines of force diverge out radially from a +ve charge and converge at a - ve charge. More correctly the lines of force are always directed from higher to lower potential.

Fig. 3.12 (2) The tangent drawn at any point on line of force gives the direction of force acting on a positive charge placed at that point.

(3) Two lines of force never intersect. If they are assumed to intersect, there will be two directions of electric field at the point of intersection, which is impossible.

(4) These lines have a tendency to contract in tension like a stretched elastic string. This actually explains attraction between opposite charges.

(5) These lines have a tendency to separate from each other in the direction perpendicular to their length. This explains repulsion between like charges.

(6) The no. of lines originating or terminating on a charge is proportional to the magnitude of charge. In rationalised MKS system $(1 / \varepsilon_0)$ electric lines are associated with unit charge. So if a body encloses a charge $q$. Total line of force associated with it (called flux) will be $\frac{q}{\varepsilon_0}$.

(7) Total lines of force may be fractional as lines of force are imaginary.

(8) Lines of force ends or strarts normally on the surface of a conductor.

(9) If there is no electric field there will be no lines of force.

(10) Lines of force per unit area normal to the area at a point represents magnitude of intensity, crouded lines represent strong field while distant lines represent weak field.

(11) Electric lines of force differ from magnetic lines of force.

(a) Electric lines of force never form closed loop while magnetic lines are always closed or extended to infinity.

(b) Electric lines of force always emerge or terminate normally on the surface of charged conductor, while magnetic lines emerge or terminate on the surface of a magnetic material at any angle.

(c) Electric lines of force do not exist inside a conductor but magnetic lines of force may exist inside magnetic material. Lines of force do not exist inside a conductor (as field inside a conductor is zero) as shown. (Electrostatic shielding)

CHECK POINT:

Which among the curves shown in Figure cannot possibly represent electrostatic field lines?

Show Answer

SOLUTION

(a) It is wrong, because electric field lines must be normal to the surface of conductor outside it.

(b) It is wrong because field lines cannot start or originate from negative charge, and also cannot end or submerge into positive charge.

(c) It is correct

(d) It is wrong because electric field lines never intersect each other.

(e) It is wrong because electric field lines cannot form closed loops.

ELECTRIC FLUX

A surface $S$ may be considered to be placed in an electric field $\vec{E}$ and imagined to be subdivided into many elementary parts, each of area $\Delta S$. The electric field over each elementary area may be assumed uniform. Then, the electric flux through an elementary part is defined as

$ \Delta \phi_E=(E \cos \theta) \Delta S=\vec{E} \cdot \Delta \vec{S} $

where $E \cos \theta$ is the normal component of $\vec{E}$ and $\Delta \vec{S}$ is a surface vector, having its magnitude equal to the area of elementary part and its direction outwards along the normal. Evidently, this electric flux may be positive, or zero, or negative, depending on the angle $\theta$ between the electric field $\vec{E}$ and normal. Then, the electric flux through the whole surface $S$ is

$ \phi_E=\Delta \phi _{E_1}+\Delta \phi _{E_3}+ $

where the summation is performed over all the elementary parts of $S$. By definition, $\phi_E$ is a scalar which is positive for the outward flux and negative for the inward flux.

For a closed surface $S$ placed in an electric field $\vec{E}$ in vacuum, we have

$ \phi_E=\frac{q}{\varepsilon_0} $

where $q$ is the total charge anywhere inside of the surface. If there is no charge inside of the surface, then $\phi_E=0$. This statement is called the ‘Gauss’s Law, or Flux Theorem’.

If a charge $+q$ is situated at the centre of a cubical surface $S$, the total electric flux emanated from the charge and passing through all the six identical faces is $\phi_E=q / \varepsilon_0$ so that the electric flux through each of the faces becomes

If the charge $+q$ is fastened at the centre of ceiling of a cubical surfaces $S$, the electric flux emanated from the charge and falling on the walls and floor is

If the charge $+q$ is placed at one corner of the cubical surface $S$, this comer may be imagined to be common to the eight identical cubical surfaces so that the electric flux through one cubical surface becomes

$ \phi_E=\frac{q}{8 \varepsilon_0} $

and then, the electric flux through each of the three faces meeting at that corner is zero and the electric flux through each of the other three faces is

ILLUSTRATION-3.7

A surface encloses an electric quadrupole. What is the total electric flux through the surface?

Show Answer

SOLUTION:

An electric quadrupole consists of the two electric dipoles, each one having the two point charges $+q,-q$ and the zero net charge. Therefore, the net charge of a quadrupole is zero, and then, according to the Gauss’s Law, the total electric flux through the surface enclosing the quadrupole given by the net charge inside of the surface divided by the permittinity $\varepsilon_0$, is zero.

ILLUSTRATION-3.8

A cylinder of radius $R$ and length $L$ is placed in a uniform electric field $\vec{E}$, parallel to it’s axis, as shown. Calculate the total electric flux through the surface of cylinder.

Show Answer

SOLUTION:

The cylinder is simply placed in the uniform electric field $\vec{E}$ which may otherwise be in any direction, and it does not have any charge enclosed inside of it. Therefore, according to the Gauss’s law, the total electric flux through the surface of cylinder must be zero.

ELECTRIC POTENTIAL

Electric potential at a point in electric field is defined to be equal to the minimum work done by an external agent in moving a unit positive charge from infinity or a reference point to that point against the electrical force of the field.

Electric potential at a point in electric field is numerically equal but opposite in sign to the work performed by electrical force to bring unit positive charge from infinity to that point.

If $W$ is the work done by external agent in bringing a positive test charge $q_0$ from infinity to a point then the potential $V$ at that point,

$ V=\frac{W _{ext}}{q_0} ; V=-\frac{.W_E \text{ (work done by electric field })}{q_0} $

Unit of potential is joule/coulomb or volt. (S.I. unit)

Potential due to Many Point Charges

The electrostatic potential due to many point charges $q_j \equiv q_1, q_2, q_3, \ldots \ldots$ at a point at the distances $r_j \equiv r_1, r_2, r_3, \ldots \ldots \ldots .$. , respectively, from the charges $q_j$ is given by the algebraic summation, i.e..

$ V=V_1+V_2+V_3+\equiv \sum_j V_j $

where $j=1,2,3, \ldots .$. . The electrostatic potential due to a point charge $q_j$ at a point at the distance $r_j$ is

$ V_j=(\frac{1}{4 \pi \varepsilon_0 K})(\frac{q_j}{r_j}) $

For example, the electrostatic potential at the origin $x=0$ due to an infinite number of positive point charges, each of magnitude, $q$, situated at the points $x=1 m, 2 m, 4 m, \ldots \ldots .$. is

$\begin{aligned} V & =\left(\frac{1}{4 \pi \varepsilon_0 K}\right)\left(\frac{q}{1}+\frac{q}{2}+\frac{q}{4}+\ldots \ldots \ldots \ldots . . . . . .\right. \\ & =\left(\begin{array}{c} q \\ 4 \pi \varepsilon_0 K \end{array}\right)\left(1+\frac{1}{2}+\frac{1}{4}+\ldots \ldots . . .\right. \\ & =\left(\begin{array}{c} q \\ 4 \pi \varepsilon_0 K \end{array}\right)\left(\begin{array}{c} 1 \\ 1-1 / 2 \end{array}\right)=2\left(\frac{q}{4 \pi \varepsilon_0 K}\right) \end{aligned}$

If the infinite number of point charges are alternately positive and negative, then

$\begin{aligned} V & =\left(\frac{q}{4 \pi \varepsilon_0 K}\right)\left(\frac{q}{1}-\frac{q}{2}+\frac{q}{4} \ldots \ldots \ldots \ldots . . . .\right. \\ & =\left(\frac{q}{4 \pi \varepsilon_0 K}\right)\left(1-\frac{1}{2}+\frac{1}{4} \ldots \ldots . .\right) \\ & =\left(\frac{q}{4 \pi \varepsilon_0 K}\right)\left(\frac{1}{1+1 / 2}\right)=\frac{2}{3}\left(\frac{q}{4 \pi \varepsilon_0 K}\right) \end{aligned}$

ELECTRIC POTENTIAL DIFFERENCE

Potential deference between two points is equal to the minimum work done in moving a unit positive test charge from one point to the other.

$ V_B-V_A=\frac{W _{A B}}{q_0} \Rightarrow W _{A B}=q_0(V_B-V_A) $

$W_E$ is the work done by the electric ficld then ;

$ \begin{aligned} & -(V_B-V_A)=\frac{W_E}{q_0} \Rightarrow W_E=-q_0(V_B-V_A) \\ & W _{\text{ext. }}=-W_E \end{aligned} $

UNDERSTANDING POTENTIAL DIFFERENCE

When the ends of a heat conductor are at different temperatures, heat energy flows from the higher temperature to the lower temperature. The flow ceases when both ends reach the same temperature. Similarly, when the ends of an electrical conductor are at different electric potentials - when there is a potential difference - charges in the conductor flow from the higher potential to the lower potential. The flow of charge persists until both ends reach the same potential. Without a potential difference, no flow of charge will occur.

To attain a sustained flow of charge in a conductor, some arrangement must be provided to maintain a difference in potential while charge flows from one end to the other. The situation is analogous to the flow of water from a higher reservoir to a lower one fig. Water will flow in a pipe that connects the reservoirs only as long as a difference in water level exists. The flow of water in the pipe, like the flow of charge in a wire, will cease when the pressures at each end are equal. (We imply this phenomenon when we say that water seeks its own level.) A continuous flow is possible if the difference in water levels - hence the difference in water pressures - is maintained with the use of a suitable pump (in Figure).

(a) Water flows from the reservoir of higher pressure to the reservoir of lower pressure. The flow will cease when the difference in pressure ceases.

(b) Water continue to flow because a difference in pressure is maintained with the pump. Fig. 3.25

A sustained electric current requires a suitable pumping device to maintain a difference in electric potential - to maintain a voltage. Chemical batteries or generators are “electrical pumps” that can maintain a steady flow of charge. These devices do work to pull negative charges apart from positive ones. In chemical batteries, this work is done by the chemical disintegration of zinc or lead in acid, and the energy stored in the chemical bonds is converted to

electric potential energy.

Fig.3.26 Each coulomb of charge that is made to flow in a circuit that connects the ends of this 1.5V. Flashlight cell is energized with 1.5J.

Generators separate charge by electromagnetic induction, a process we will describe in the next chapter. The work that is done (by whatever means) in separating the opposite charges is available at the terminals of the battery or generator. This energy per charge provides the difference in potential (voltage) that provides the “electrical pressure”. to move electrons through a circuit joined to those terminals.

A common automobile battery will provide an electrical pressure of 12 volts to a circuit connected across its terminals. Then 12 joules of energy are supplied to each coulomb of charge that is made to flow in the circuit.

EQUIPOTENTIAL SURFACES

A surface of constant, or uniform, potential $V$ is called the equipotential surface, e.g., the surface of a charged conductor carrying no electric currents, any spherical surface about a point charge as centre, any plane surface drawn parallel to and between the plates of a charged parallel-plate capacitor and so on. For a given charge distribution, the equipotential surfaces and eletric-field

lines form an orthogonal system, i.e., they are mutually perpendicular. The equipotential surfaces do not intersect each other and they are closer in the region of stronger electric field and widely separated in the region of weaker electric field.

A volume of constant, or uniform, potential $V$ is called the equipotential volume, e.g., the volume of a charged conductor, carrying no electric currents.

CHECK POINT:

Two charges $2 \mu$ C and $-2 \mu$ C are placed at points $A$ and $B, 6 cm$ apart.

(a) Identify an equipotential surface of the system.

(b) What is the direction of the electric field at every point on this surface?

Show Answer

SOLUTION

(a) Since it is an electric dipole, so a plane normal to $AB$ and passing through its mid-point has zero potential everywhere.

(b) Normal to the plane in the direction $AB$.

ILLUSTRATION-3.9

Ten positively charged particles have the charges $1 \times 10^{-8} C, 4 \times 10^{-8} C, 9 \times 10^{-8} C, 16 \times 10^{-8} C \ldots . . . ., 100 \times 10^{-8} C$ and are situated on the $x$-axis at the points $x=0.1 m, 0.2 m, 0.3 m$, ., $1.0 m$, respectively. Calculate the electric potential at the origin $O$.

Show Answer

SOLUTION:

The electric potential at the origin $O$ due to the given ten positively-charged particles is

$ \begin{aligned} & V=(9 \times 10^{9}) \times[\frac{1 \times 10^{-8}}{0.1}+\frac{4 \times 10^{-8}}{0.2}+\frac{9 \times 10^{-8}}{0.3}+\ldots \ldots . .+\frac{100 \times 10^{-8}}{10}] \\ & =90 \times[10+20+30+\ldots \ldots \ldots .+100] \\ & =900 \times[1+2+3+\ldots \ldots \ldots . .+10] \\ & =900 \times 55 \\ & =4.95 \times 10^{4} \text{ volt. } \end{aligned} $

ILLUSTRATION-3.10

A uniform electric field of $10 N / C$ exists in the vertically downward direction. Find the increase in the electric potential as we go up through a height of $0.5 m$.

Show Answer

SOLUTION:

The increase in the electric potential $(V_B-V_A)$ as we go up from the point $A$ to the point $B$ through a height of $0.5 m$ is equal to the work done in displacing a charge of $+1 C$ from $A$ to $B$ against the electric field. Therefore, we have

$ \Delta V=(V_B-V_A)=+10 \frac{N}{C} \times 1 C \times 0.5 m=5 V $

ILLUSTRATION-3.11

A radioactive source, in the form of a metallic sphere of the radius $0.01 m$, emits the $\beta$ particles at the rate of $5 \times 10^{10}$ particles/s. The source is electrically insulated. How long will it take for it’s electric potential to the emitted $\beta$ particles escape the source?

Show Answer

SOLUTION:

Say, the electric potential of the source is raised to the $2 V$ in the time $t$ seconds. In this time, the total number of $\beta$ particles emitted from the radioactive source is

$ n=5 \times 10^{10} \times t $

and the number of $\beta$ particles escaping the source is

$ n^{\prime}=\frac{40}{100} \times 5 \times 10^{10} \times t $

The fotal charge of these $\beta$ particles is

$ q=-e \times n^{\prime}=-e \times \frac{40}{100} \times 5 \times 10^{10} \times t C $

and hence, the total charge which appears on the radioactive source is

$ q^{\prime}=-q=+e \frac{40}{100} \times 5 \times 10^{10} \times t C $

Due to this charge, the source, being spherical and of radius $R=0.01 m$, is raised to the electric potential of $2 V$. Therefore,

$ \begin{aligned} 2 & =(\frac{1}{4 \pi \varepsilon_0})(\frac{q^{\prime}}{R}) \\ 2 & =(9 \times 10^{9}) \frac{(e \times \frac{40}{100} \times 5 \times 10^{10} \times t)}{0.01} \\ \Rightarrow \quad t & =\frac{2 \times 0.01 \times 100}{9 \times 10^{9} \times 1.6 \times 10^{-19} \times 40 \times 5 \times 10^{10}}=\frac{1}{1440} \text{ second. } \end{aligned} $

ELECTROSTATIC POTENTIAL ENERGY

The electrostatic potential energy of a point charge $q$ placed at a point in the electric field $\vec{E}$ is given by

$ U=q V $

i.e., it is the work done in displacing the point charge $q$ from the reference point which is arbitrarily chosen at infinity and $U$ is arbitraily taken zero there, to that point. The electrostatic potential energy gain $(U_B-U_A)$ in displacing the charge $q$ from the point $A$ to the point $B$ in the electric field $\vec{E}$ is equal to the work done in displacing the charge $q$ from $A$ and $B$. Therefore,

$ (U_B-U_A)=q(V_B-V_A) $

where $(V_B-V_A)$ is the electrostatic potential difference between the points $A$ and $B$. This gain may be positive, or zero, or negative, depending on the charge and the electrostatic potential difference. For example, when an electron of charge-e moves from a lower to higher potential, it loses its electrostatic potential energy. The loss, or gain, of electrostatic potential energy is nommally expressed in the unit, i.e.,

$ 1 eV \equiv 1.6 \times 10^{-19} J $

The electrostatic potential energy of a system of two point charge $q_1$ and $q_2$, separated by a distance $r _{12}$, is

$ U=(\frac{1}{4 \pi \varepsilon_0 K})(\frac{q_1 q_2}{r _{12}}) $

which is defined as the amount of work done in assembling the two point charges from the infinite separation between them. The electrostatic potential energy of a system of three point charges $q_1, q_2$ and $q_3$, having the separation $r _{12}$ between $q_1, q_2$. separation $r _{23}$ between $q_2, q_3$ and separation $r _{31}$ between $q_3, q_1$, becomes

$ U=(\begin{matrix} 1 \\ 4 \pi \varepsilon_0 K \end{matrix} )(\begin{matrix} q_1 q_2+q_2 q_3+q_3 q_1 \\ r _{12} \\ r _{23} \end{matrix} ) $

which is equal to the amount of work done in assembling the system of three point charges from the intinite separations between them.

CHECK POINT:

A balloon may easily be charged to several thousand volts. Does that mean it has several thousand joules of energy? Explain.

Show Answer

SOLUTION:

No, it doesn’t mean that the balloon has several thousand joules of energy always. It can have several thousand joules of energy only if the balloon acquires high charge.

ILLUSTRATION-3. 12

Three point charges $q_1, q_2$ and $q_3$ of $0.1 C$ each are placed at the corners of an equilateral triangle of the side $1 m$. If this system is supplied energy at the rate of $1 kW$, then how much time will be required to move one of the charges to the mid point of the line joining the others?

Show Answer

SOLUTION:

The three charges are at the corners of an equilateral triangle, as shown.

The total electrostatic potential energy of the system is

$ \begin{aligned} U_1 & =(\frac{1}{4 \pi \varepsilon_o})(\frac{q_1 q_2}{r _{12}}+\frac{q_2 q_3}{r _{23}}+\frac{q_3 q_1}{r _{31}}) \\ & =(9 oe 10^{9})(\frac{0.1 \times 0.1}{1}+\frac{0.1 \times 0.1}{1}+\frac{0.1 \times 0.1}{1})=2.7 \times 10^{8} J \end{aligned} $

When one of the charges, say, $q_1$, is moved to the mid point of the line joining the other two charges, as shown,

the total electrostatic potential energy of the system will be

$ \begin{aligned} U_2 & =(9 \times 10^{9})(\frac{0.1 \times 0.1}{0.5}+\frac{0.1 \times 0.1}{0.5}+\frac{0.1 \times 0.1}{1}) \\ & =4.5 \times 10^{8} J \end{aligned} $

Then, the total energy supplied to the system becomes

$ \Delta U=U_2-U_1=(4.5-2.7) \times 10^{8}=1.8 \times 10^{8} J $

As the power supplied to the system is $1 kW \equiv 1000 W$, we have the time taken in supplying the energy $\Delta U$, i.e.,

$t=\frac{1.8 \times 10^8}{1000}=1.8 \times 10^5 s=50 \mathrm{~h}$.

ILLUSTRATION-3.13

Two identical point charges, $+q$ and $+q$ are placed on the $x$-axis at the points $x=a$ and $x=-a$, respectively. A third point charge $Q$ is placed at the origin $x=0$. Prove that the change in the lectrostatic potential energy of $Q$ is approximately,

$ \Delta U=(\frac{1}{4 \pi x_0})(\frac{2 Q q d^{2}}{a^{3}}) $

when it is displaced by a distance $d«$ a along the $x$-axis.

Show Answer

SOLUTION:

Before the displacement of the point charge $Q$, it’s electrostatic potential energy is

$ \begin{aligned} & U_1=(\frac{1}{4 \pi \varepsilon_0})(\frac{Q q}{a}+\frac{Q q}{a})=(\begin{matrix} 1 \\ 4 \pi \varepsilon_0 \end{matrix} )(\begin{matrix} 2 Q q \\ a \end{matrix} ) \end{aligned} $

When $Q$ is displaced by a distance $d«a$ in the $+x$ direction, then it’s electrostatic potential energy becomes

$ \begin{aligned} & U_2=(\frac{1}{4 \pi \varepsilon_0})[\frac{Q q}{(a+d)}+\frac{Q}{(a-d)}]=(\begin{matrix} 1 \\ 4 \pi \varepsilon_0 \end{matrix} ) \begin{bmatrix} 2 a Q q \\ (a^{2}-d^{2}) \end{bmatrix} \\ & =(\frac{1}{4 \pi \varepsilon_0})(\frac{2 Q q}{a})(1-\frac{d^{2}}{a^{2}})^{-1} \approx(\frac{1}{4 \pi \varepsilon_0})(\begin{matrix} 2 Q q \\ a \end{matrix} )(1+\frac{d^{2}}{a^{2}}) \end{aligned} $

Now, the change, i.e., increase, in the electrostatic potential energy of $Q$ becomes

$ \begin{aligned} \Delta U & =U_2-U_1 \\ & \approx(\frac{1}{4 \pi \varepsilon_0})(\frac{2 Q q}{a})(\frac{d^{2}}{a^{2}}) \approx(\frac{1}{4 \pi \varepsilon_0})(\begin{matrix} 2 Q q d^{2} \\ a^{3} \end{matrix} ) \end{aligned} $

ELECTRIC DIPOLE

An electric dipole is a system of two unlike point charges $-q,+q$, having a negligibly small separation $l$. It has no net charge and it is characterised only by its electric dipole moment, i.e.

$ \vec{p}=q \vec{l} $

where $q$ is the magnitude of the point charges, $\vec{l}$ is the vector separation directed from the negative point charge to the positive point charge. Evidently, $\vec{p}$ has the magnitude ( $q l$ ) and the direction from the negative point charge to the positive point charge.

The torque $\tau$ on a dipole in uniform electric field as shown in figure is given by, $\tau=q E \times 2 \ell \sin \theta=\vec{p} \times \vec{E}$ So $\tau$ is maximum, when dipole is $\perp$ to field & minimum $(=0)$ when dipole is parallel or antiparallel to field. If $\vec{p}=p_x \hat{i}+p_y \hat{j}+p_z \hat{k}$ and $\vec{E}=E_x \hat{i}+E_y \hat{j}+E_z \hat{k}$

Then $\vec{\tau}=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ p_x & p_y & p_z \\ E_x & E_y & E_z\end{array}\right|$

The work done in rotating the dipole from equilibrium through an angle $\mathrm{d} \theta$ is given by

$\mathrm{dW}=\tau \mathrm{d} \theta=\mathrm{pE} \sin \theta \mathrm{d} \theta$ and from $\theta_1 \rightarrow \theta_2$, $W=\int_{\theta_1}^{\theta_2} d W=\int_{\theta_1}^{\theta_2} p E \sin \theta d \theta$

If $\theta_1=0$ i.e equilibrium position, then

$ W=\int_0^{\theta_2} pE \sin \theta d \theta=pE(1-\cos \theta_2) $

Workdone in rotating an electric dipole in uniform electric field from $\theta_1$ to $\theta_2$ is $W=pE(\cos \theta_1-\cos \theta_2)$

Time period of a dipole in uniform electric field is

$ T=2 \pi \sqrt{\frac{l}{p E}} $

where I = Moment of Inertia of the dipole about the axis of rotation.

Potential energy of an electric dipole in an electric field is.

$ U=-\vec{p} \cdot \vec{E} \text{ i.e: } U=-p E \cos \theta $

where $\theta$ is the angle between $\vec{E}$ and $\vec{p}$.

We can also write

Electric field due to an electric dipole

$ U=p_x E_x+p_y E_y+p_z E_z $

(a) Axial line (end-on position)

$\overrightarrow{p}$ and $\vec{E}$ are parallel

(b) Equatorial line (Broad - side on potential)

$ E _{Ax}=\frac{1}{4 \pi \hat I _0} \cdot \frac{2 px}{(x^{2}-1^{2})^{2}} \text{ when } x \gg 1 $

$ \begin{gathered} \overrightarrow{E} \to p \\ -q \frac{x}{\ell+q} \\ E _{eq}=\frac{1}{4 \pi \epsilon_0} \cdot \frac{p}{(x^{2}+\ell^{2})^{3 / 2}} \text{ where } x \gg l \end{gathered} $

$\vec{p}$ and $\vec{E}$ are anti parallel i.e when $x \gg>l$.

(c) At any point (for first dipole)

$ E _{A x}=2 E _{e q} $

$ E=\frac{1}{4 \pi \epsilon_0} \frac{p}{x^{3}} \sqrt{3 \cos ^{2} \theta+1} ; \tan \beta=\frac{1}{2} \tan \theta $

(d) Electric field intensity due to a point charge varies inversely as cube of the distance and in case of quadrupole it varies inversely as the fourth power of distance from the quadrupole.

ILLUSTRATION-3.14

Calculate the electric intensity due to a dipole of length $10 cm$ and having a charge of $500 \mu C$ at a point on the axis distant $20 cm$ from one of the charges in air.

Show Answer

SOLUTION:

The electric intensity on the axial line of the dipole is given by

$ E _{\text{axial }}=\frac{1}{4 \pi \varepsilon_o} \times \frac{2 rp}{[r^{2}-(\frac{a}{2})^{2}]^{2}} $

Here $a=10 cm$ or $a / 2=5 \times 10^{-2} m, r=(20+5) cm=25 \times 10^{-2} m, p=q \times a=(500 \times 10^{-6}) \times(10 \times 10^{-2})=5 \times 10^{-5} C-m$

$ E _{\text{axial }}=(9 \times 10^{9}) \times \frac{2 \times(25 \times 10^{-2}) \times(5 \times 10^{-5})}{10^{-8}[(25)^{2}-(5)^{2}]^{2}}=3.25 \times 10^{7} \cdot N^{-}-C $

CAPACITORS AND CAPACITANCE

A capacitor or condenser consists of two conductors separated by an insulator or dielectric. The conductors carry equal and opposite charges $\pm Q$. It is an electrical device that store electric charge. The conductors used in capacitor are called plates. Whatever their shape may be (square, circular, rectangular, spherical or cylindrical, all are called plates). A single conductor acts as a conductor, with the assumption that the other plate is at infinity and earthed. If some how the potential is lowered, to bring it once again to the same potential more charge can be given to it, so the capacitance of the conductor can be increased.

The capacitance of a conductor is gently increased when an earth connected conductor is placed near it.

Capacitance

When charge is given to an isolated body its potential increases

i.e., $q \propto V \Rightarrow q=C V \Rightarrow C=\frac{q}{V}$

if, $\quad V=1 \Rightarrow C=q$

(1) The ability to store charge is called capacitance of the capacitor. Capacitance of a body is numerically equal to the change required to raise the potential by unit.

(2) Units : SI unit of $C \Rightarrow C /$ volt $=Farad(F)$

C.G.S. unit $\to 1 F=\frac{C}{\text{ volt }}=\frac{3 \times 10^{9} \text{ stat } C}{1 / 300 \text{ stat volt }}$

$\Rightarrow 1 F=900 \times 10^{9}=9 \times 10^{11}$ stat Farad.

Dimensions : $[M^{-1} L^{-2} T^{4} A^{2}]$

In case of spherical conductor

$V=\frac{q}{4 \pi \varepsilon_0 R} \Rightarrow \frac{q}{V}=C=4 \pi \varepsilon_0 R$

(b)

If we assume the earth to be a conducting sphere of radius $6400 km$ its capacity will be $C=4 \pi \varepsilon_0 R=711 \mu F$.

(3) The capacity of the body is independent of the charge given to the body or its potential raised and depends on its shape and size only. $\quad C \propto q^{0}, C \propto V^{0}$

Capacitance depends on :

(a) Area of the conductor: If the surface area of a conductor is increased keeping charge q const. Its potential decreases and hence the capacity increases $C=\frac{q}{V}$

(b) The presence of other conductor in the vicinity of the charged conductor. The presence of other conductor near a charged conductor decreases the potential of the charged conductor, there by the capacity increases.

(c) The medium around the charged conductor: If the capacity of the conductor is $C$ when air is the medium atound it, then the capacity becomes $K C$ when another medium of dielectric constant $K$ is around it. This is because of the reason that potential becomes $V / K$.

140 Static Electricity I Physics I

ENERGY STORED IN A CAPACITOR

When a conductor is charged its potential changes from 0 to $V$ and work is done against repulsion betw een charge stared on the conductor and charge coming from the charging body. This work is stored as electrostatic potential energy ( $U$ ).

$ U=\frac{q^{2}}{2 C}=\frac{1}{2} q V=\frac{1}{2} C V^{2} . $

The energy is stored in the electric field between the plates of the capacitor

The energy stared per unit valume in the electric field between the plates is called energy dersidy

$u=\frac{1}{2} \varepsilon_0 E^{2}=\frac{1}{2} \frac{\sigma^{2}}{\varepsilon_0}$.

WORK DONE BY BATTERY

Work done by battery is given by

$\mathbf{W}=$ Charge that flow through battery $\times$ battery EMF

Whenever there is a charging of capacitor, work done by battery, in part is stored as electrostaric enargy in termeen carwicior plates and remaining is dissipated as heat due to charge flow through connecting wires.

In general, heat produced is given by

$H=$ Work done by battery - Energy stored in capacitor

TYPES OF CAPACITORS

Capacitors are mainly of three types:

(1) Parallel plate capacitor

(2) Spherical capacitor

(3) Cylindrical capacitor

(1) Parallel plate capacitor (PPC)

A PPC consists of two equal parallel metal plates facing each other and sparand by 2 dielectric $(\varepsilon_r)$. The plates may be square, rectangular or circular in shape.

Let us consider a parallel-plate capacitor, having a charge $q$ and the surface charge density $\sigma$ on ins two platses with a separation $d$ and area $A$ of each face. The electric field between the two plates is

$E_0=\frac{\sigma}{\varepsilon_0}=\frac{q}{\varepsilon_0 A}$

and electric potential difference across them is

$ V=(V_A-V_B)=E_0 d=\frac{q d}{E_0 A} $

Then, the capacitance of capacitor becomes

$ C=\frac{q}{V}=\frac{E_0 A}{d} $

In case of PCC

(i) Capacity is directly proportional to effective over-laping area of the plate $C \propto A$

(ii) Capacity is inversely proportional to the distance between the plate fie. $C \times 1 / d$.

(iii) Capacity is independent of charge given, potential raised or nature of metal and thichness of plates

(2) Spherical capacitor

(i) Outer Spherical Conductor Grounded

Let us consider a spherical capacitor, having the two spherical conductors $A$ and $B$ of the radii $a$ and $b$, respectively, with the conductor $B$, grounded. When a charge $+q$ is given to the inner spherical conductor $A$, a charge $-q$ is induced on the inner surface and a charge $+q$ is induced on the outer surface of the outer conductor $B$. The induced charge $+q$ flows to the ground and only, the induced charge $-q$ remains on the inner surface of the outer conductor $B$.

The electric potential of the inner spherical conductor $A$ due to the two charges, $+q$ on $A$ and $-q$ on $B$, becomes

$ V_A=(\frac{1}{4 \pi \varepsilon_0 K})(\frac{+q}{a}+\frac{-q}{b})=(\begin{matrix} q \\ 4 \pi \varepsilon_0 K \end{matrix} )(\begin{matrix} b-a \\ a b \end{matrix} ) $

The electric potential of the outer spherical conductor $B$ due to the two charges becomes

$ V_B=(\begin{matrix} 1 \\ 4 \pi \varepsilon_0 K \end{matrix} )(\frac{+q}{a}+\frac{-q}{b})=0 $

Then, the capacitance of the spherical capacitor, having the two spherical conductors $A$ and $B$, with the outer conductor $B$ grounded, is given by

$ C=\frac{q}{(V_A-V_B)}=\begin{gathered} q \\ V_A \end{gathered}=(4 \pi \varepsilon_0 K)(\begin{matrix} a b \\ b-a \end{matrix} ) $

When the outer conductor $B$ assumes the infinitely large size, i.e., the radius $b$ tends to infinity, only the inner spherical conductor remains on the scene and its capacitance is given by

$ C=(4 \pi \varepsilon_0 K)=(\frac{a}{1-\frac{a}{b}})=(4 \pi \varepsilon_0 K) a $

The capacitance of our earth which is assumed to be spherical conductor of mean radius of $6400 km$, is about

$ C=(4 \pi \varepsilon_0) a=\frac{6400 \times 1000}{9 \times 10^{9}} F \approx 711 \mu F $

(ii)Inner Spherical Conductor Grounded

Let us, now, consider a spherical capacitor, having the two spherical conductors $A$ and $B$ of the radii $a$ and $b$, respectively, with the inner spherical conductor $A$ grounded. In order to calculate the capacitance of such a spherical capacitor, we may introduce a third grounded and concentric spherical conductor $D$ of the radius $d>b>a$. Then; the system of three conductors is equivalent to a parallel combination of the two spherical capacitors of capacitances $C_1, C_2$, as shown in the figure. The capacitance of the system of three conductors becomes

$ \begin{aligned} C & =C_1+C_2 \\ & =(4 \pi \varepsilon_0 K)(\frac{a b}{b-a})+(4 \pi \varepsilon_0 K)(\frac{b d}{d-b}) \\ & =(4 \pi \varepsilon_0 K)(\begin{matrix} a b+b d \\ b-a \end{matrix} ) \\ & =(4 \pi \varepsilon_0 K)(\frac{a b}{b-a}+\frac{b d}{1-b / d}) \end{aligned} $

(b)

Fig. 3.35

When the conductor $D$ assumes the infinity large size, i.e., the radius $d$ tends to infinity, only the spherical conductors $A, B$ remain on the scene and the capacitance of system, then, becomes

$ C=(4 \pi \varepsilon_0 K)(\frac{a b}{b-a}+b)=(4 \pi \varepsilon_0 K)(\frac{b^{2}}{b-a}) $

which is more than the capacitance of the system, with the outer spherical conductor $B$ grounded, by an amount $(4 \pi \varepsilon_0 K) b$, i.e., the capacitance of the outer spherical conductor $B$ only.

ILLUSTRATION-3.15

The radius of circular plates of a parallel-plate capacitor is $r$. What is the distance between the plates, if it’s capacitance, with the air as dielectric, is equal to that of an isolated sphere of radius $r^{\prime}$ in air?

Show Answer

SOLUTION:

The capacitance of parallel-plate capacitor, with the air as dielectric, is

$ C=K(\frac{\varepsilon_0 A}{d})=K(\frac{\varepsilon_0 \pi r^{2}}{d}) $

where $A=\pi r^{2}$ is the area of one face of each plate of capacitor, $d$ is the separation between the plates of capacitor and $K$ is the dielectric constant of air. Equating this capacitance to the capactance of an isolated sphere of radius $r^{\prime}$ in air, i.e.,

$ C^{\prime}=(4 \pi \varepsilon_0 K) r^{\prime} $

we have, $\quad K(\frac{\varepsilon_0 \pi r^{2}}{d})=(4 \pi \varepsilon_0 K) r^{\prime}$

$ \Rightarrow \quad d=(\frac{r^{2}}{4 r^{\prime}}) $

(3) Cylindrical capacitor

A cylindrical capacitor consists of the two coaxial cylindrical conductors $A$ and $B$ of the radii $a$ and $b$, respectively, with the outer cylindrical conductor $B$ grounded. If the inner cylindrical conductor $A$ is given a charge per unit length of $+\lambda$, a charge per unit length of $-\lambda$ is induced on the inner surface of the outer cylindrical conductor $B$.

The electric field between the cylindrical conductors at the distance $r$ from the common axis is

$ E=\frac{\lambda}{2 \pi \varepsilon_0 K r} \text{ for }(a<r<b) $

which gives the potential for the two cylindrical conductors

$ \begin{aligned} & V_A=-(\frac{\lambda}{2 \pi \varepsilon_0 K}) \log _{e} a+\text{ constant } \\ & V_B=-(\frac{\lambda}{2 \pi \varepsilon_0 K}) \log _{e} b+\text{ constant } \end{aligned} $

The electric potential difference between the two cylindrical conductors becomes

$ V=(V_A-V_B)=(\frac{\lambda}{2 \pi \varepsilon_0 K}) \log _{e}(\frac{b}{a}) $

and then, the capacitance of the cylindrical capacitor becomes

$ C=\frac{q}{V}=\frac{\lambda l}{V}=\frac{(2 \pi \varepsilon_0 K) l}{\log _{e}(b / a)}=\frac{(2 \pi \varepsilon_0 K) l}{2.3030 \times \log _{10}(b / a)} $

where the charge on the capacitor is $q=\lambda l, l$ being the length of capacitor

To increase the capacity of a cylindrical capacitor

(i) $K$ or $\varepsilon_r$ symbol be more

(ii) b/a should be less i.e. $b \approx a$

(iii) Length of the cylinder (L) should be more.

ILLUSTRATION-3.16

Two cylindrical capacitors, each of the capacitance of $2.2 \mu F$, are kept with their outer cylinders in contact and their inner cylinders connected through a conducting wire, as shown. What is the equivalent capacitance of system as a whole?

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SOLUTION:

The given system is a parallel combination of two cylindrical capacitors, each of capacitance of $2.2 \mu F$, as shown.

Therefore, the equivalent capacitance of system as a whole is

$ C=2.2+2.2=4.4 \mu F $

CHECK POINT:

What meaning would you give to the capacitance of a single conductor?

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SOLUTION:

It means thata single conductor is a capacitor whose other plate can be considered to be at infinity.

COMBINATION OF CAPACITORS

Series Combination

Let us consider a number of capacitors, connected in series, as shown.

They have the same charge $q$ but the different electric potential differences $V_1, V_2, V_3$. capacitances $C_1, C_2, C_3, \ldots \ldots .$. . The total electric potential difference applied across the combination is $V=V_1+V_2+V_3+$

$ \begin{aligned} & =\frac{q}{C_1}+\frac{q}{C_2}+\frac{q}{C_3}+\ldots \ldots \ldots . . \\ & =q(\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}+\ldots \ldots \ldots . . . . .. \end{aligned} $

Therefore, the equivalent capacitance $C$ of the series combination is given by

$ \frac{1}{C}=\frac{V}{q}=\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}+ $

Evidently, therefore, the equivalent capacitance $C$ is smaller than each one of the capacitances of combination.

Parallel Combination

We, now, consider a number of capacitors, connected in parallel, as shown. They have the same electric potential difference $V$ across them, but the different charges $q_1, q_2, q_3 \ldots \ldots .$. on them due to their different capacitances $C_1, C_2, C_3$

The total charge of the combination is

$ \begin{aligned} q & =q_1+q_2+q_3+\ldots \ldots \ldots . . . . . . . . . . . \end{aligned} $

from which we have the equivalent capacitance of the combination, i.e.,

$ C=\frac{q}{V}=C_1+C_2+C_3 \ldots \ldots $

Evidently, therefore, the equivalent capacitance $C$ is more than each one of the capacitances of combination.

ILLUSTRATION-3.17

A circuit is shown in figure.

$ \text{ Fig. } 3.41 $

Find the charge on the condenser having a capacity of $5 \mu F$.

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SOLUTION:

Potential difference between $A$ and $B=6$ volts. The condensers $2 \mu F$ and $5 \mu F$ are in parallel. Their effective capacitance, $C=2$+5=7 $\mu F$.

The capacitance between $A$ and $B$ is given by $C^{\prime}=\frac{C \times 3}{C+3}=\frac{7 \times 3}{7+3}=\frac{21}{10} \mu F$

Total charge $Q=CV=\frac{21}{10} \times 6=\frac{63}{5} \mu C$

Total potential difference across $3 \mu F$ is

$V_1=\frac{Q}{3}=\frac{63}{5} \times \frac{1}{3}=\frac{21}{5}$ volts

Hence the common potential difference across the condensers in parallel is

$v_2=6-\frac{21}{5}=\frac{9}{5} v$

So, the charge on $5 \mu F$ condenser is

$Q=5 \times V_2=5 \times \frac{9}{5}=9 \mu C$

ILLUSTRATION-3.18

In the given circuit, consisting of the four capacitors and a battery, the potential difference across the $7 \mu F$ capacitors is $6 V$. Then, calculate the potential differences across the remaining three capacitors and the battery emf.

Fig. 3.42

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SOLUTION:

The capacitors of $7 \mu F$ and $3 \mu F$ are in series and therefore, they have the same charge on them and the potential differences across them are in a ratio equal to the inverse ratio of capacitances. Therefore, the potential difference across the $3 \mu F$ capacitor is

$ =\frac{6 V \times 7 \mu F}{3 \mu F}=14 V $

The potential difference across the parallel combination of $3.9 \mu F$ capacitor with the series combination of $7 \mu F$ and $3 \mu F$ capacitor is, therefore, $20 V$. The total capacitance of this combination is

$ =3.9+\frac{7 \times 3}{7+3}=6 \mu F $

which is in series with the capacitor of $12 \mu F$ and therefore, they have same charge and the potential differences are in a ratio equal to the inverse ratio of capacitances. The potential difference across the $12 \mu F$ capacitor becomes

and then, the battery emf becomes

$ =\frac{20 V \times 6 \mu F}{12 \mu F}=10 V $

$ =20+10=30 V \text{. } $

LOSS OF ENERGY ON JOINING TWO CHARGED CONDUCTORS OR CAPACITORS

When the two charged conductors, having the capacitances $C_1$ and $C_2$, respectively, and the potentials $V_1$ and $V_2$, respectively, are connected together by the conducting wires, the charge flows from the higher potential, say, $V_1$ to the lower potential, say, $V_2$ till, finally, the potentials of the two conductors are equalised. Then, the common potential of the two conductors will be

$ V=\frac{(q_1+q_2)}{C_1+C_2}=\frac{(C_1 V_1+C_2 V_2)}{(C_1+C_2)} $

The same thing, as outlined above, happens, when the two charged capacitors are connected, with the plates of the like polarities together, by the conducting wires. Before connecting together, the total energy of the charged conductors, or capacitors, is

$U_1= _2^{1}(C_1 V_1^{2})+ _2^{1}(C_2 V_2^{2})$

After connecting together, the total energy of charged conductors, or capacitors, will be

$ U_2=\frac{1}{2}(C_1+C_2) V^{2}=\frac{(C_1 V_1+C_2 V_2)}{2(C_1+C_2)} $

Then, the loss of energy becomes

$ \Delta U=U_1-U_2=\frac{1}{2}(\begin{matrix} C_1 C_2 \\ (C_1+C_2) \end{matrix} )(V_1-V_2)^{2} $

which is mainly as the heat and also as the light and sound due to the sparking taking place. This loss is zero, when $V_1=V_2$

$\Rightarrow \frac{q_1}{C_1}=\frac{q_2}{C_2} \Rightarrow \frac{q_1}{q_2}=\frac{C_1}{C_2}$

i.e., the charges of the two conductors, or capacitors, before joining together are in the ratio of their capacitances. After connecting together, the charges on the two conductors, or capacitors, will also be in the ratio of their capacitancces, i.e.,

$ \frac{q_1^{\prime}}{q_2^{\prime}}=\frac{C_1 V}{C_2 V}=\frac{C_1}{C_2} $

ILLUSTRATION-3.19

A $4 \mu F$ capacitor is charged by a $200 V$ supply. It is then disconnected from the supply and then, connected to another uncharged $2 \mu F$ capacitor. How much electrostatic potential energy of the first capacitor is lost in the form of heat and other radiation?

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SOLUTION:

The electrostatic potential energy of the first capacitor of $4 \mu F$ capacitance, lost in the form of heat and other radiation, when it is charged by a $200 V$ supply and connected to another unchanged capacitor of $2 \mu F$ capacitances after disconnection from the supply, is

$ \Delta U=\frac{1}{2}(\frac{C_1 C_2}{C_1+C_2})(V_1 \sim V_2)^{2}=\frac{1}{2}(\frac{4 \times 10^{-6} \times 2 \times 10^{-6}}{4 \times 10^{-6}+2 \times 10^{-6}})(200-0)^{2}=2.67 \times 10^{-2} J $

MISCELLANEOUS

SOLVED EXAMPLES

1. .Two protons in a molecule is separated by $3 \times 10^{-10} m$. Find the electrostatic force exerted by one proton on the other..

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Solution.

According to Coulomb’s law, the electrostatic force $F$ between two charges $q_1$ and $q_2$, which are seperated by distance $r$ is

Here, $q_1=q_2=1.6 \times 10^{-19} C, r=3 \times 10^{-10} m$

$ F=\frac{1}{4 \pi \varepsilon_0} \times \frac{q_1 q_2}{r^{2}} $

So,

$F=9 \times 10^{9} \frac{(1.6 \times 10^{-19})^{2}}{9 \times 10^{-20}}=2.56 \times 10^{-9} C \quad$ (Repulsive)

2. Two negative charges of unit magnitude each and a positive charge $q$ are placed along a straight line. At what position and for what value of $q$ will the system be in equilibrium? Check whether it is stable, unstable or neutral equilibrium?

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Solution . Let the charge $+q$ be held at a distance $x_1$ from unit negative charge at $A$, and at a distance $x_2$ from unit negative charge at $B$.

For equilibrium of $q, \frac{q(-1)}{4 \pi \varepsilon_o x_1^{2}}=\frac{q(-1)}{4 \pi \varepsilon_o x_2^{2}}$ $\therefore x_1=x_2$ i.e. $q$ must be equidistant from’ $A$ and $B$.

For equilibrium of unit negative charge at $B$.

Force on $B$ due to charge at $A+$ force on $B$ due to $q=0$

$\frac{(-1)(-1)}{4 \pi \varepsilon_o(x_1+x_2)^{2}}+\frac{q(-1)}{4 \pi \varepsilon_o x_2^{2}}=0$

$\Rightarrow \quad \frac{1}{4 \pi \varepsilon_o(2 x_2)^{2}}=\frac{-(-q)}{4 \pi \varepsilon_o x_2^{2}} \quad(\because x_1=x_2)$

$q=\frac{1}{4}$ i.e. $\frac{1}{4}$ th of the magnitude of either unit charge.

Stability : If $q$ is displaced slightly towards $A$, force of attraction due to $A$ exceeds the force of attraction due to $B$. Therefore, $q$ will get displaced further towards $A$. Hence the equilibrium of $q$ is unstable.

However, if $q$ is displaced in a direction $\perp$ to $A$, net force would bring $q$ back to its normal position. Therefore, the equilibrium will be stable.

3. A $10 mF$ capacitor is charged to a potential difference of $50 V$ and is connected to another uncharged capacitor in parallel. Now the common potential difference becomes 20 volt. What is the capacitance of second capacitor is?

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Solution . $q_1=10 \times 50=500 \mu C, C_1=10 \mu F, C_2=$ ?, $q_2=0$

As $V=\frac{q_1+q_2}{C_1+C_2}$

$\therefore C_1+C_2=\frac{q_1+q_2}{V}=\frac{500+0}{20}=25 \mu F$

$C_2=25-C_1=25-10=15 \mu F$

4. Electric force between two point charges $q$ and $Q$ at rest is $F$. Now if a charge $-q$ is placed next to $q$ what will be the (a) force on $Q$ due to $q(b)$ total force on $Q$ ?

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Solution .

(a) As electric force between two body interaction, i.e., force between two particles, is independent of presence or absence of other particles, the force between $Q$ and $q$ will remain unchanged, i.e., $F$.

(b) An electric force is proportional to the magnitude of charges, total force on $Q$ will be given by :

$\frac{F^{\prime}}{F}=\frac{Q q^{\prime}}{Q q}=\frac{q^{\prime}}{q}=\frac{0}{q}=0 \quad[.$ as $.q^{\prime}=q+(-q)=0]$

i.e., The resultant force on $Q$ will be zero.

5. Calculate the electric field strength required to just support a water drop of mass $10^{-7} kg$ and having charge $1.6 \times 10^{-19} C$.

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Solution . Here, $m=10^{-7} kg, q=1.6 \times 10^{-19} C$

Step 1 : Let $E$ be the electric field strength required to support the water drop.

Force acting on the water drop due to electric field $E$ is

Weight of drop acting downward,

$ F=q E=1.6 \times 10^{-19} E $

$ W=mg=10^{-7} \times 9.8 \text{ newton. } $

Step 2 : Drop will be supported if $F$ and $W$ are equal and opposite.

i.e., $1.6 \times 10^{-19} E=9.8 \times 10^{-7}$

or

$ E=\frac{9.8 \times 10^{-7}}{1.6 \times 10^{-19}}=6.125 \times 10^{12} NC^{-1} $

6. The following figure shows a surface $S$ which is enclosing $-2 q$ charge. The charge $+q$ is kept outside the surface $S$. Calculate the net outward/inward flux from the surfaces.

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Solution . According to Gauss law, the net flux is

Fig. 3.45

$\phi=\frac{1}{\varepsilon_o} \times$ net charge enclosed by closed surface $=\frac{-2 q}{\varepsilon_0}$

(because $+q$ is out side the surface $S$, so net flux due to $+q$ thought $S$ is zero)

7. In which Figure, the electric flux is maximum?

Fig. 3.46

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Solution . According to Gauss Law, the electric flux linked with a closed surface depends only on net charge enclosed by that surface. It does not depend on the shape and size of that closed surface. Hence electric flux linked in above three figures are same i.e., $\phi_A=\phi_B=\phi_C$

8. A charge $q$ is enclosed in a cube. What is the electric flux associated with one of the faces of cube

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Solution . According to Gauss theorem the total electric flux

$\phi=\frac{1}{\varepsilon_0} \times$ total enclosed charge $=\frac{1}{\varepsilon_0} \times q$.

Since cube has six faces. Hence electric flux linked with each face $=(1 / 6 \phi)=q / 6 \varepsilon_0$.

9. A charge $Q \mu C$ is placed at the centre of a cube. What will be the flux coming out from any surface will be

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Solution . According to Gauss’s law

$\phi($ flux from all six surfaces of cube $)=\frac{Q \mu C}{\varepsilon_0}=\frac{Q}{\varepsilon_o} \times 10^{-6}$

So flux from any one of surface of cube $=\frac{Q}{6 \varepsilon_o} \times 10^{-6}$

10. $n$ small drops of same size are charged to $V$ volt each. If they coalese to form a single large drop, then what will be its potential?

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Solution . As Volume remains constant, therefore,

$ \begin{gathered} \frac{4}{3} \pi R^{2}=n \times \frac{4}{3} \pi r^{3} \quad \therefore R=n^{1 / 3} r . \\ \text{ New potential }=V^{\prime}=\begin{matrix} n q \\ 4 \pi \varepsilon_0 R \end{matrix} =\begin{matrix} n q \\ 4 \pi \varepsilon_0(n^{1 / 3} r) \end{matrix} =n^{2 / 3} \quad \begin{matrix} q \\ 4 \pi \varepsilon_o r \end{matrix} =n^{2 / 3} V . \end{gathered} $

1 EXERCISE

Fill in the Blanks:

DIRECTIONS : Complete the following statements with an appropriate word / term to be filled in the blank space(s).

1. Electric is of two types viz. positive and negative charge.

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Answer: (charge)

2. Proton is said to be charged and electron is said to be charged

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Answer: (positively, Negatively)

3. Like charges

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Answer: (repuls, attract)

4. A proton repels a and unlike charges

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Answer: (proton/electron)

5. The force of attraction or repulsion between two is given by

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Answer: (charges, Coulomb’s law)

6. Charge can neither be nor

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Answer: (created, destroyed)

7. The charge from one body can be to another body but the total charge of a system remains

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Answer: (transferred, constant)

8. The static electricity deals with electric charges at and their

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Answer: (rest, effects)

9. Charge flows from a body at to a body at

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Answer: (higher potential, lower potential)

10. To bring a unit positive charge from to a point in an electric field, some work has to done which is called

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Answer: infinity, electric potential

11. An electric dipole is a system of two point charges

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Answer: urilike

12. The capacitance of conductor depends on the and of conductor and the surrounding medium.

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Answer: shape, size

13. A system of two insulated from each other, is called a capacitor.

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Answer: Conductors

14. The equivalent capacitance $C$ of the series combination is given by $\frac{1}{C}=$

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Answer: $\left(\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}+\ldots \ldots \ldots . . ..\right)$

15. The equivalent capacitance $C$ of the parallel combination is given by $\mathrm{C}=$

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Answer: $\left(C_1+C_2+C_3 \ldots \ldots \ldots . . . . ..\right)$

True /False:

DIRECTIONS : Read the following statements and write your answer as true or false.

1. The electric charge is of two types viz, positive and negative charge.

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Answer: True

2. Like charges attracts to each other and unlike charges repels to each other.

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Answer: False

3. A proton repels an electron.

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Answer: False

4. A nutron repels a proton.

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Answer: False

5. Charge can neither be created nor destroyed.

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Answer: True

6. Proton is said to be charged positively and electron is said to be charged negatively.

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Answer: True

7. The branch of electricity which deals with the properties -of electrified bodies due to a moving charge is called electrostatics.

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Answer: False

8. In nature charge $(Q)$ always occurs as an integral multiple of electronic charge ’ $e$ ’ i.e. $Q=$ ne, where $n=1,2,3$,

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Answer: True

9. Charge can be in fraction of e, i.e. 1.7e, 0.8e, or -1.3e.

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Answer: False

10. The electronic charge ’ $e$ ’ is the maximum possible charge,

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Answer: False

11. The device used to detect the presences of charge and to Identify the nature of charge on a body, is called an electroscop.

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Answer: True

12. An electric dipole is a pair of equal and same charges.

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Answer: False

13. The mutual electrostatic force between two point charges $q_1$ and $q_2$ is inversely proportional to the product of charges $q, q_2$.

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Answer: False

14. Electric field due to a point charge $q$ has a magnitude $\frac{|q|}{4 \pi \varepsilon_0 r^3} \text {. }$

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Answer: False

15. Potential at a point is the force due to electric dipole.

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Answer: False

16. For capacitors in the series combination, the total capacitance $C$ is given by $\mathrm{C}=\mathrm{C}_1+\mathrm{C}_2+\mathrm{C}_3+$

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Answer: False

Match The Following:

DIRECTIONS: Each question contains statements given in tavo columns which have to be matched. Statements (A, B, C, D) in column I have to be matched with statements $(p, q, r, s)$ in column II.

1.

$ \begin{array}{ll} \text{Column I} & \text{Column II} \\ \text{(A) Like charges repels and unlike attracts} & \text{(p) Dr. William Gilbert} \\ \text{(B) Numerical value of force between two charges} & \text{(q) Thomos Brown} \\ \text{(C) Methods of charging} & \text{(r) Coulomb by (torsion balance)} \\ \text{(D) Amount of induced charge} & \text{(s) Faraday ice pail exp.} \\ \end{array} $

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Answer: $(\mathrm{A}) \rightarrow \mathrm{p}$; (B) $\rightarrow \mathrm{r}$; (C) $\rightarrow \mathrm{q}$; (D) $\rightarrow \mathrm{s}$

2.

$ \begin{array}{ll} \text{Column I} & \text{Column II} \\ \text{(A) e/m of electron} & \text{(p) J.J Thomson} \\ \text{(B) Charge and mass (indirectly) of electron and quanta of charge.} & \text{(q) R.A. Millikan (by oil drop exp.)} \\ \text{(C) Concept of line of force} & \text{(r) M. Faraday} \\ \text{(D) Highest common factor method} & \text{(s) Max Plank} \\ \end{array} $

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Answer: (A) $\rightarrow \mathrm{p}$; (B) $\rightarrow \mathrm{q}$; (C) $\rightarrow \mathrm{r}$; (D) $\rightarrow \mathrm{s}$

3. Two point charges $Q$ and $-Q / 4$ are separated by $L$ Column I

$ \begin{array}{lll} \text{Column I} & & \text{Column II} \\ \text{(A) Zero potential} & &\text {(p) $L / 3 $ on right side of charge $-Q / 4$ } \\ \text{(B) Zero electric field} & & \text{(q) $L / 5$ on left side of charge $-Q / 5$} \\ \text{(C) Infinite potential} & & \text{(r) $L$ on right side of charge $-Q / 4$} \\ \text{(D) Infinite electric field} & & \text{(s) Near charge Q.} \\ \end{array} $

Show Answer

Answer: $\quad(A) \rightarrow p, q ;(B) \rightarrow r ;(C) \rightarrow s ;(D) \rightarrow s$

Very Short Answer Questions:

DIRECTIONS : Give answer in one word or one sentence.

1. Write the physical quantity which has its unit coulomb/ volt? Is it a vector or a scalar quantity?

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Answer: The quantity is capacitance $(C=\frac{Q}{V})$. It is a scalar quantity.

2. Write down the relation between electric field and potential at a point.

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Answer: $E=-\frac{dV}{dr}$ i.e. electric field at a point is the negative of the electric potential gradient at that point.

3. Is electrostatic potential necessarily zero at a point where electric field strength is zero?

Illustrate your answer.

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Answer: No, it is not necessary.

$ \because \quad E=-\frac{d V}{d \dot{r}} $

$\therefore$ If $V$ is constant, $E$ will be zero.

Ex: The electric field inside a hollow spherical conductor is zero but potential is not zero.

4. What is the workdone in moving a charge of $50 nC$ between two points on an equipotential surface?

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Answer: Workdone to move the charge $50 nC$ on the equipotential surface is zero as the potential at every point is same.

5. Draw an equipotential surface in a uniform electric field.

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Answer:

It is a plane surface perpendicular to the electric field.

6. What is the ratio of electric field intensities at any two points between the plates of a parallel plate capacitor?

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Answer: Required ratio is $1: 1$, since the electric field is uniform between the plates of a capacitor.

7. The work done in moving a charge of $3 C$ between two points is $6 J$. What is the potential difference between the points?

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Answer: $V=\frac{W}{q}=\frac{6}{3}=2 V$.

8. In a parallel plate capacitor, the capacitance increases from $4 \mu F$ to $80 \mu F$, on introducing a dielectric medium between the plates. What is the dielectric constant of the medium?

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Answer: $C=\frac{\varepsilon_0 A}{d}$ and $C^{\prime}=\frac{k \varepsilon_0 A}{d}$

$\therefore \frac{C^{\prime}}{C}=k \Rightarrow k=\frac{80}{4}=20$

9. The equivalent capacitance of the combination between $A$ and $B$ in the given figure is $15 \mu F$. Calculate the capacitance of capacitor $C$.

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Answer: $C$ and $20 \mu F$ are in series. $C _{\text{eq }}=15 \mu F$

$\therefore \frac{1}{15}=\frac{1}{20}+\frac{1}{C} \Rightarrow \frac{1}{C}=\frac{1}{15}-\frac{1}{20}$

$\Rightarrow \frac{1}{C}=\frac{4-3}{60}=\frac{1}{60} \Rightarrow C=60 \mu F$.

10. Two large conducting spheres carrying charges $Q_1$ and $Q_2$ are brought close to each other. Is the magnitude of electrostatic force between them exactly given by $Q_1 Q_2 / 4 \pi \varepsilon_0 r^{2}$, where $r$ is the distance between their centres?

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Answer: No, because charge distributions on the spheres will not be uniform.

11. If Coulomb’s law involved $1 / r^{1}$ dependence (instead of $1 / r^{2}$ ), would Gauss’s law be still true?

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Answer: No.

12. A small test charge is released at rest at a point in an electrostatic field configuration. Will it travel along the field line passing through that point?

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Answer: Not necessarily. (True only if the field line is a straight line.) The field line gives the direction of acceleration, not that of velocity, in general.

13. We know that electric field is discontinuous across the surface of a charged conductor. Is electric potential also discontinuous there?

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Answer: No, potential is continuous.

14. What meaning would you give to the capacitance of a single conductor?

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Answer: A single conductor is a capacitor with one of the ‘plates’ at infinity.

15. What meaning would you give to the capacitance of a single conductor?

Show Answer

Answer: A water molecule has permanent dipole moment. However, detailed explanation of the value of dielectric constant requires microscopic theory and is beyond the scope of the book.

16. Name the physical quantity whose SI unit is (a) newton/ coulomb, (b) joule/coulomb.

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Answer: Newton/coulomb is the SI unit of electric field strength.

(b) Joule/coulomb is the SI unit of electric potentials.

$\therefore$ Force will decrease as $k$ increases.

17. What orientation of an electric dipole in a uniform electric field corresponds to its stable equilibrium?

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Answer: If the dipole is placed along the direction of the field, its equilibrium is stable $(\theta=0^{\circ})$.

18. Calculate the no. of electrons in $1 C$ charge.

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Answer: $1 e=1.6 \times 10^{-19} C$

$\therefore$ No. of electrons $=\frac{1}{1.6 \times 10^{-19}}=6.25 \times 10^{18}$

19. Draw lines of force to represent a uniform electric field.

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Answer:

20. Which particles are responsible for the flow of current in conductors?

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Answer: Electrons.

21. What is the SI unit of electric potential?

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Answer: In SI, electric potential is measured in volt $(V)$

22. Write the mathematical expression for electric potential.

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Answer: Mathematically, electric potential is written as

$ V=\frac{W}{q} $

Short Answer Questions:

DIRECTIONS : Give answer in 2-3 sentences.

1. Two point electric charges of unknown magnitude and sign are placed at a distance ’ $d$ ’ apart. The electric field intensity is zero at a point, not between the charges but on the line joining them. Write two essential conditions for this to happen.

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Answer: (a) The two charges are of opposite sign so that force will be attractive in nature.

(b) The two charges have different magnitudes - the charge of smaller magnitude will be nearer to the point where the total field intensity is zero.

2. Electric charge is distributed uniformly on the surface of a spherical rubber balloon. Show how the value of. electric intensity vary (i) on the surface (ii) inside and (iii) outside?

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Answer: (i) On the surface, $E=$ constant

(ii) Inside the surface, $E=\mathbf{0}$.

(iii) Out side the surface, $E \propto \frac{1}{r^{2}}$.

3. A surface encloses an electric quadrupole. What is the total electric flux through the surface?

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Answer: An electric quadrupole consists of the two electric dipoles, each one having the two point charges $+q,-q$ and the zero net charge. Therefore, the net charge of a quadrupole is zero and, then, according to the Gauss’s Law, the total electric flux through the surface enclosing the quadrupole, given by the net charge inside of the surface divided by the permittivity $\varepsilon_0$, is zero.

4. What is the force between two small charged spheres having charges of $2 \times 10^{-7} C$ and $3 \times 10^{-7}$ placed $30 cm$ apart in air?

Show Answer

Answer: Given, $q_1=2 \times 10^{-7} C, q_2=3 \times 10^{-7} C, r=30 cm=0.3$ $m, F=$ ?

By Formula, $F=\frac{1}{4 \pi \varepsilon_0} \frac{q_1 q_2}{r^{2}}$

$=\frac{9 \times 10^{9} \times 2 \times 10^{-7} \times 3 \times 10^{-7}}{(0.3)^{2}}=6 \times 10^{-3} N$ (repulsive)

5. When a glass rod is rubbed with a silk cloth, charges appear on both. A similar phenomenon is observed with many other pairs of bodies. Explain how this observation is consistent with the law of conservation of charge.

Show Answer

Answer: Charge is neither created nor destroyed. It is merely transferred from one body to another. Electrons are transferred from glass to silk, so glass has positive charge and silk has negative charge.

6. (a) An electrostatic field line is a continuous curve. That is, a field line cannot have sudden breaks. Why not?

(b) Explain why two field lines never cross each other at any point?

Show Answer

Answer: (a) They start from a positive charge and end at a negative charge. They are continuous, because force is continuous. They do not have sudden breaks, otherwise a moving test charge will have to take jumps.

(b) Two lines of force do not intersect each other. If they intersect at a point, there will be two directions of field at that point. Since it is impossible, hence they don’t intersect.

7. Three capacitors each of capacitance $9 pF$ are connected in series. (a) What is the total capacitance of the combination? (b) What is the potential difference across each capacitor, if the combination is connected to a $120 V$ supply?

Show Answer

Answer: (A) $\frac{1}{C}=\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}=\frac{3}{9 \times 10^{-12}}$ $\Rightarrow C=3 \times 10^{-12}=PF$

(B) $V=V_1+V_2+V_3=120 \Rightarrow \frac{q}{C_1}+\frac{q}{C_2}+\frac{q}{C_3}=120$

$\Rightarrow C_1=C_2=C_3=9 \times 10^{-12}$

$\frac{3 q}{9 \times 10^{-12}}=120 \Rightarrow q=360 \times 10^{-12} C$.

$V=\frac{q}{C}=\frac{360 \times 10^{-12}}{9 \times 10^{-12}}=40 V$.

8. Calculate the value of $C$ in the following combination if the equivalent capacitance between $P$ and $Q$ is $30 \mu F$.

Show Answer

Answer:

The $3,20 \mu F$ capacitors are in parallel.

$\therefore \quad 20+20+20=60 \mu F \Rightarrow \frac{1}{C}+\frac{1}{60}=\frac{1}{30}$ $\quad \Rightarrow C=60 \mu F$.

9. What is an equipotential surface? Show that the electric field is always directed perpendicular to an equipotential surface.

Show Answer

Answer: An equipotential surface is that at every point of which electric potential is same. Consider two points A and B on the equipotential surface.

By definition, potential difference between two points $A$ and $B=$ work done in carrying a unit positive charge from $A$ to $B$.

$\Rightarrow V_A-V_B=W _{AB}=\overrightarrow{E} \cdot d \ell \quad But_A=V_B$

$\therefore \overrightarrow{E} d l=0$

$\Rightarrow$ Edl $\cos \theta=0 \Rightarrow \cos \theta=0 \Rightarrow \theta=90^{\circ}$

$\therefore \overrightarrow{E} \perp d l$

$\therefore \quad$ Electric field $(\overrightarrow{E}$ ) is directed perpendicular to the equipotential surface.

10. Equipotential surfaces are perpendicular to field lines. Why?

Show Answer

Answer: On an equipotential surface, no work is done in moving a charge from one point to another. Therefore, the component of electric field intensity along the equipotential surface is zero. So equipotential surfaces are perpendicular to field lines.

11. Distinguish between electric potential and potential energy.

Show Answer

Answer: Electric potential at a point is the amount of workdone in moving a unit positive charge from infinity to that point.

Potential energy is the amount of workdone in carrying the total charge from infinity to that point against the electrostatic forces.

12. Is it possible for a metal sphere of $1 cm$ radius to hold a charge of $1 C$ ?

Show Answer

Answer: We have $Q=CV$ or $V=\frac{Q}{C}$

$\Rightarrow \frac{Q}{4 \pi \varepsilon_0 r}=\frac{9 \times 10^{9} \times 1}{10^{-2}}=9 \times 10^{11}$ Volt.

So, it is not possible for a metallic sphere to hold a charge of $1 C$ because the potential is so high that electric breakdown of air will occur and the entire charge will leak out.

13. What is the relation between dielectric constant and electric susceptibility?

Show Answer

Answer: The relation between dielectric constant and electrical susceptibility of a material is

$ K=1+\chi $

The electrical susceptibility describes the electrical behaviour of a dielectric. It has different values for different dielectrics. For vacuum $\chi=0$.

14. An electric dipole is free to move in a uniform electric field. Explain its motion when it is placed (i) parallel to the field and (ii) perpendicular to the field.

Show Answer

Answer: The torque acting on a dipole placed in a uniform electric field at an angle $\theta$ is $t=\pi E \sin \theta$

(i) When the dipole is parallel to the field, $\overline{p}$ is along $\overrightarrow{E}$ $\therefore \theta=0$

$\therefore t=pE \sin \theta=0$

(ii) When the dipole is perpendicular to the field is perpendicular to $\overline{E}$

$\therefore \theta=90^{\circ}$

$\therefore$ Torque acting on it is maximum.

$\therefore t _{\max }=pE$

15. Two capacitors of capacitances $C_1=3 \mu F$ and $C_2=6 \mu F$ arranged in series are connected in parallel with a 3rd capacitor $C_3=4 \mu F$. The total arrangement is connected to a $6 V$ battery. Calculate the total energy stored is the capacitors.

Show Answer

Answer:

$\frac{1}{C_s}=\frac{1}{3}+\frac{1}{6} \Rightarrow C_s=2 \mu F \Rightarrow C_p=C_s+C_3=2+4=6 \mu F$

Energy stored $=\frac{1}{2} CV^{2}=\frac{1}{2} \times 6 \times 10^{-6} \times 6^{2}=108 \times 10^{-4} J$.

Long Answer Questions:

DIRECTIONS : Give answer in four to five sentences.

1. (a) Explain the meaning of the statement ’electric charge of a body ‘is quantised’.

(b) Why can one ignore quantisation of electric charge when dealing with macroscopic i.e., large scale charge?

Show Answer

Answer: (a) Quantization of electric charge. It is now a well known fact that all charges occurring in nature are positive or negative integral multiples of a basic unit of electric charge which we take as the magnitude of the charge on an electron. We use symbol $e$ for the amount of charge on an electron. Hence charge on an electron is $-e$ and that on a proton happens to be $+e$, while charge on a neutron is zero. Any charged body will have $\pm$ ne charge, where $n$ is an integer. This fact is called the quantization of electric charge.

(b) At the macroscopic level one deals with charges that are enormous compared to the magnitude of charge $e$. Since $e=1.6 \times 10^{-19} C$, a charge of magnitude, say, $1 \mu C$ contains something like $10^{13}$ times the electronic charge. At this scale, the fact that charge can increase or decrease only in units of $e$ is not very different from saying that charge can take continuous values. Thus, at the macroscopic level, the quantisation of charge has no practical consequence and can be ignored.

2. A capacitor is charged to potential $V_1$. The power supply is then disconnected and the capacitor is then connected in parallel to another capacitor of potential $V_2$

Derive an expression for the common potential of the combination of capacitor.

Show Answer

Answer: Let $C_1$ and $C_2$ are the capacitances, $q_1$ and $q_2$ are the charges, $V_1$ and $V_2$ are potential of the capacitors respectively.

$ \therefore q_1=C_1 V_1 \quad \text{ and } q_2=C_2 V_2 $

Before sharing, total charge $=q_1+q_2=C_1 v_1+C_2 v_2$

When the capacitors are joined by a wire, charge will flow from higher to lower potential till both the potentials are equal. This equal potential is called common potential $(V)$

If $q_1^{\prime}$ and $q_2^{\prime}$ are charges on $C_1$ and $C_2$ after redistribution of charges, then

$ q_1^{\prime}=C_1 V \text{ and } q_2^{\prime}=C_2 V $

$\because$ Total charge after connecting them together remains same as before.

$ \begin{aligned} & q _{=} q_1^{\prime}+q_2^{\prime}=C_1 V+C_2 V=(C_1+C_2) V \\ \therefore \quad & (C_1+C_2) V=C_1 V_1+C_2 V_2 \\ & V=\frac{C_1 V_1+C_2 V_2}{C_1+C_2} \end{aligned} $

3. Derive the expression for the capacitance of a parallel plate capacitor filled with dielectric.

Show Answer

Answer:

Let $A$ be the plate area of the capacitor. $d$ is the distance between the plates.

$q$ be the charge on each plate of the capacitor. $k$ is the dielectric constant of the medium between the plates of the capacitor. $\sigma$ be the surface charge density

$\therefore \quad$ Electric field between the plates $=E=\frac{\sigma}{k \varepsilon_0} ;$ As the

potential is same, $V=Ed=\frac{\sigma}{k \varepsilon_0} d$

Capacitance $=C=\frac{q}{V}=\frac{q}{\frac{\sigma d}{k \varepsilon_0}}$

$\therefore C=\frac{\sigma A}{\frac{\sigma d}{k \varepsilon_0}}=\frac{k \varepsilon_0 A}{d} \quad[\because q=\sigma A]$

4. How is Coulomb’s law similar to Newton’s law of gravitation? How is it different?

Show Answer

Answer: Coulomb’s law states that, for two charged objects that are much smaller than the distance between them, the force between them varies directly as the product of their charges and inversely as the square of the separation distance and it acts along a straight line from one charge to the other. It can be expressed as where, $d$ is the distance between the charged particles, $q_1$ and $q_2$ represent the charges on the two particles, and $k$ is the proportionality constant. Newton’s law of gravitation can be expressed as

$ F=\frac{k q_1 q_2}{d^{2}} $

where, $F$ is the force of attraction between two objects of masses $m_1$ and $m_2$ separated by a distance $d$.

$ \frac{G m_1 m_2}{d^{2}} $

From equations (1) and (2), it seems that Coulomb’s law and Newton’s law of gravitation are similar in the way that both follow the inverse square law and take the same form. But the two laws are different in the way that Coulombian force can be attractive or repulsive according as the nature of charges but gravitational force between two masses is always attractive in nature.

5. How does the magnitude of electric force compare with the charge between a pair of charged particles when they are brought to half their original distance of separation? To One-quarter their original distance? To four times their original distance? (What law guides your answers?)

Show Answer

Answer: According to inverse-square law, the electric force between two charges is inversely proportional to the square of the distance of their separation. Therefore, when the charges are brought to half their original distance of separation, the electric force between them increases by four times.

If the separation distance is made one-quarter their original distance, the force increases by sixteen times.

Similarly, the force between the charges decreases by sixteen times when the separation distance between them is made four times their original distance.

(The inverse square law guides our answer).

2 EXERCISE

Multiple Choice Questions:

DIRECTIONS : This section contains multiple choice questions. Each question has 4 choices (a), (b), (c) and (d) out of which ONLY ONE is correct.

1. An electric dipole, consisting of two opposite charges of $2 \times 10^{-6}$ Ceach separated by a distance $3 cm$ is placed in an electric field of $2 \times 10^{5} N / C$. Torque acting on the dipole is

(a) $12 \times 10^{-1} N-m$

(b) $12 \times 10^{-2} N-m$

(c) $12 \times 10^{-3} N-m$

(d) $12 \times 10^{-4} N-m$

Show Answer

Answer: (c) Charges $(q)=2 \times 10^{-6} C$, Distance

(d) $=3 cm=3 \times 10^{-2} m$ and electric field (E)

$=2 \times 10^{5} N / C$. Torque $(\tau)=$ q.d.

$E=(2 \times 10^{-6}) \times(3 \times 10^{-2}) \times(2 \times 10^{5})$

$=12 \times 10^{-3} N-m$.

2. If the potential of a capacitor having capacity $6 \mu F$ is increased from $10 V$ to $20 V$, then increase in its energy will be

(a) $4 \times 10^{-4} J$

(b) $4 \times 10^{-6} J$

(c) $9 \times 10^{-4} J$

(d) $12 \times 10^{-6} J$

Show Answer

Answer: (c) Capacitance of capacitor $(C)=6 \mu F=6 \times 10^{-6} F$; Initial potential $(V_1)=10 V$ and final potential $(V_2)=20 V$.

The increase in energy $(\Delta U)$

$=\frac{1}{2} C(V_2^{2}-V_1^{2})$

$=\frac{1}{2} \times(6 \times 10^{-6}) \times[(20)^{2}-(10)^{2}]$

$=(3 \times 10^{-6}) \times 300=9 \times 10^{-4} J$.

3. Intensity of an electric field (E) depends on distance $r$, due to a dipole, is related as

(a) $E \propto \frac{1}{r}$

(b) $E \propto \frac{1}{r^{2}}$.

(c) $E \propto \frac{1}{r^{3}}$

(d) $E \propto \frac{1}{r^{4}}$

Show Answer

Answer: (c) Intensity of electric field due to a Dipole

$E=\frac{p}{4 \pi \varepsilon_0 r^{3}} \sqrt{3 \cos ^{2} \theta+1} \Rightarrow E \propto \frac{1}{r^{3}}$

4. A charge $q$ is placed at the centre of the line joining two exactly equal positive charges $Q$. The system of three charges will be in equilibrium, if $q$ is equal to

(a) $-\frac{Q}{4}$

(b) $+Q$

(c) $-Q$

(d) $\frac{Q}{2}$

Show Answer

Answer: (a) At equilibrium, net force is zero

$\therefore \frac{1}{4 \pi \varepsilon_0} \times \frac{Qq}{(r / 2)^{2}}+\frac{1}{4 \pi \varepsilon_0} \times \frac{Q^{2}}{r^{2}}=0$

or $4 q=-Q$ or $q=-\frac{Q}{4}$.

5. The formation of a dipole is due to two equal and unlike point charges placed at a

(a) short distance

(b) long distance

(c) above each other

(d) none of these

Show Answer

Answer: (a) Dipole is formed when two equal and unlike charges are placed at a short distance.

6. A parallel plate condenser with oil between the plates (dielectric constant of oil $K=2$ ) has a capacitance $C$. If the oil is removed, then capacitance of the capacitor becomes

(a) $\sqrt{2} C$

(b) $2 C$

(c) $\frac{C}{\sqrt{2}}$

(d) $\frac{C}{2}$

Show Answer

Answer: (d) When oil is placed between space of plates

$C=\frac{2 A \varepsilon_0}{d} \ldots(1)[\because C=\frac{K A \varepsilon_0}{d}.$, if $.K=2]$

When oil is removed $C^{\prime}=\frac{A \varepsilon_0}{d} \ldots$ (2)

On compari both equations, we get $C=C / 2$

7. When air is replaced by a dielectric medium of force constant $K$, the maximum force of attraction between two charges, separated by a distance

(a) decreases $K$-times

(b) increases $K$-times

(c) remains unchanged

(d) becomes $\frac{1}{K^{2}}$ times

Show Answer

Answer: (a) In air, $F _{\text{air }}=\frac{1}{4 \pi \varepsilon_0} \frac{q_1 q_2}{r^{2}}$

In medium, $F_m=\frac{1}{4 \pi \varepsilon_0} \frac{q_1 q_2}{Kr^{2}}$

$\therefore \frac{F_m}{F _{\text{air }}}=\frac{1}{K} \Rightarrow F_m=\frac{F _{\text{air }}}{K}$ (decreases $K$-times)

8. A capacitor is charged to store an energy $U$. The charging battery is disconnected. An identical capacitor is now connected to the first capacitor in parallel. The energy in each of the capacitors is

(a) $U / 2$

(b) $3 U / 2$

(c) $U$

(d) $U / 4$

Show Answer

Answer: (a) In 1st case when capacitor $C$ attached with battery charged with the energy.

$U_1=U$ (stored energy on capacitor).

In Ind case after disconnect of battery similar capacitor is artached in parallel with Ist capacitor then $C _{eq}=C=2 C$.

Now, $\frac{U_1}{U_2}=\frac{\frac{1}{2} \frac{q^{2}}{C}}{\frac{1}{2} \frac{q^{2}}{C}}=\frac{C^{\prime}}{C}=\frac{2 C}{C} \quad(\because C^{\prime}=2 C)$

$U_2=\frac{U}{2}$

9. A charge $Q$ is placed at the comer of a cube. The electric flux through all the six faces of the cube is

(a) $Q / 3 \varepsilon_0$

(b) $Q / 6 \varepsilon_0$

(c) $Q / 8 \varepsilon_0$

(d) $Q / \varepsilon_0$

Show Answer

Answer: (d) According to Gauss’s theorem,

electric flux through a closed surface $=\frac{q}{\varepsilon_0}$

where $q$ is charge enclosed by the surface.

10. In a parallel plate capacitor, the distance between the plates is $d$ and potential difference across the plates is $V$. Energy stored per unit volume between the plates of capacitor is

(a) $\frac{Q^{2}}{2 V^{2}}$

(b) $\frac{1}{2} \varepsilon_0 \frac{v^{2}}{d^{2}}$

(c) $\frac{1}{2} \frac{V^{2}}{\varepsilon_0 d^{2}}$

(d) $\frac{1}{2} \varepsilon_0 \frac{V^{2}}{d}$

Show Answer

Answer: (b) Energy stored per unit volume

$=\frac{1}{2} \varepsilon_0^{\prime} E^{2}=\frac{1}{2} \varepsilon_0(\frac{V}{d})^{2}=\frac{1}{2} \varepsilon_0 \frac{V^{2}}{d^{2}}(\because E=\frac{V}{d})$

11. Energy stored in a capacitor is

(a) $\frac{1}{2} Q V$

(b) $Q V$

(c) $\frac{1}{Q V}$

(d) $\frac{2}{Q V}$

Show Answer

Answer: (a) Energy stored in capacitor

$ \begin{aligned} & =\frac{1}{2} CV^{2}=\frac{1}{2} Q \cdot V^{2} \quad(Q=CV) \\ & =\frac{1}{2} QV \end{aligned} $

12. Three capacitors each of capacity $4 \mu F$ are to be connected in such a way that the effective capacitance is $6 \mu F$. This can be done by

(a) connecting two in parallel and one in series

(b) connecting all of them in series

(c) connecting them in parallel

(d) connecting two in series and one in parallel

Show Answer

Answer: (d) For series, $C^{\prime}=\frac{C_1 \times C_2}{C_1+C_2}=\frac{4 \times 4}{4+4}=2 \mu F$

For parallel, $C _{eq}=C^{\prime}+C_3=2+4=6 \mu F$

13. When a body is charged by induction, then the body

(a) becomes neutral

(b) does not lose any charge

(c) loses whole of the charge on it

(d) loses part of the charge on it

Show Answer

Answer: (b) Charging by induction involves transfer of charges from one part to the other of the body. No loss of charge is involved.

14. If a body is positively charged, then it has

(a) excess of electrons

(b) excess of protons

(c) deficiency of electrons

(d) deficiency of neutrons

Show Answer

Answer: (c) Positive charge is due to deficiency of electrons.

15. On charging by conduction, mass of a body may

(a) increase

(b) decreases

(c) increase or decrease

(d) none

Show Answer

Answer: (c) On charging by conduction, body may gain mass, if it acquires negative charge. It may lose mass, if it acquires positive charge.

16. Coulomb’s law is true for

(a) atomic distances $(=10^{-11} m)$

(b) nuclear distances $(=10^{-15} m)$

(c) charged as well as uncharged particles

(d) all the distances

Show Answer

Answer: (d) Coulomb’s law is true for all distances small and large. Hence it is called a long range force.

17. Two charges are placed a certain distance apart. A metallic sheet is placed between them. What will happen to the force between the charges?

(a) increase

(b) decrease

(c) remains unchanged

(d) either ’ $a$ ’ or ’ $b$ '

Show Answer

Answer: (b) $F_m=\frac{F_0}{K}$. For metals $K=\infty$, therefore, $F_m \to 0$, in the thickness of sheet. Hence force between the charges will decrease.

18. Which of the following is best insulator?

(a) Carbon

(c) Graphite

(b) Paper

(d) Ebonite

Show Answer

Answer: (d) Ebonite is the best insulator.

19. Two spheres $A$ and $B$ of exactly same mass are given equal positive and negative charges respectively. Their masses after charging

(a) remains unaffected

(c) mass of $A<$ mass of $B$

(b) mass of $A>$ mass of $B$

(d) nothing can be said

Show Answer

Answer: (c) A loses electrons and B gains electrons. Therefore, mass of $A<$ mass of $B$.

20. What happens when some charge is placed on a soap bubble?

(a) its radius decreases

(b) its radius increases

(c) the bubble collapses

(d) none of the above

Show Answer

Answer: (b) The radius of soap bubble increases because of outward force acting on the bubble due to charging.

21. Quantisation of charge implies

(a) charge cannot be destroyed

(b) charge exists on particles

(c) there is a minimum permissible charge on a particle

(d) charge, which is a fraction of a coulomb is not possible.

Show Answer

Answer: (d) $q= \pm$ ne shows that minimum value of $q= \pm l e$. where $e=1.6 \times 10^{-19}$

Coulomb $=$ Charge of one electron

22. Two charges are placed a certain distance apart in air. If a glass slab is introduced between them, the force between them will

(a) increase

(b) decrease

(c) remains the same

(d) be zero

Show Answer

Answer: (b) $F_g=\frac{1 q_1 q_2}{4 \pi \epsilon_0 r^{2}}=\frac{F_v}{K}$

As $K$ (for glass) $>1 . \quad \therefore F_g<F_v$

23. A charge $q_1$ exerts some force on a second charge $q_2$. If third charge $q_3$ is brought near, the force of $q_1$ exerted on $q_2$

(a) decreases

(b) increases

(c) remains unchanged

(d) increases if $q_3$ is of same sign as $q_1$ and decreases if $q_3$ is of opposite sign.

Show Answer

Answer: (c) The force of $q_1$ or $q_2$ is not affiected by $q_3$. However. resultant force on $q_2$ will be affected.

24. A positively charged rod is brought near an uncharged conductor. If the rod is then suddenly withdrawn, the charge left on the conductor will be

(a) positive

(b) negative

(c) zero

(d) not sure

Show Answer

Answer: (c) Positively charged rod induces negative charge on the side of the sphere closer to the rod and an equal positive charge on the side away from the rod. When the rod is withdrawn, the negative changes move to neutralise the equal positive change. The charge left is zera

25. The ratio of electric force between two electrons to two protons separated by the same distance in air is

(a) $10^{0}$

(b) $10^{6}$

(c) $10^{4}$

(d) none

Show Answer

Answer: (a) $\frac{F _{e, e}}{F _{p, p}}=10^{0}=1$, because forre depends only on charges, distance and nature of medium.

26. The dielectric constant of a metal is

(a) $\infty$

(b) 0

(c) 1

(d) none

Show Answer

Answer: (a) Metal is a good conductor. So, its $\mathrm{K}=\infty$.

27. A metallic particle having no net charge is placed near a finite metal plate carrying a positive charge. The electric force on the particle will be

(a) towards the plate

(b) away from the plate

(c) parallel to the plate

(d) zero

Show Answer

Answer: (d)

28. If a body is charged by rubbing it, its weight

(a) remains precisely constant

(b) increases slightly

(c) decreases slightly

(d) may increase slightly or may decrease slightly

Show Answer

Answer: (d) The weight can be increased slightly, if it acquire negative charge & weight can be decreased slightly, if it acquires positive charge.

29. An electric field can deflect

(a) neutrons

(b) X-rays

(c) $\gamma$-rays

(d) $\alpha$-particles

Show Answer

Answer: (d) An electric field can deflect only a charged particle.

30. A hollow sphere of charge does not have electric field at

(a) outer point

(b) interior point

(c) beyond $2 m$

(d) beyond $100 m$

Show Answer

Answer: (b) $\mathrm{E}=0$, at any point inside the sphere.

31. If one penetrates a uniformly charged solid sphere, the electric field $E$

(a) increases

(b) decreases

(c) is zero at all points

(d) remains same as at the surface

Show Answer

Answer: (b) E decreases as we move from the surface of the sphere to its centre.

32. If one penetrates a uniformly charged spherical cloud, electric field strength

(a) decreases directly as the distance from the centre

(b) increases directly as the distance from the centre

(c) remains constant

(d) none of the above

Show Answer

Answer: (a) A spherical cloud behaves as a solid sphere. Therefore, E decreases directly with the decreasing distance from the centre.

33. Electric lines of force about a negative point charge are

(a) circular anticloc

(b) circular clockwise

Show Answer

Answer: (c) For a single negative point charge, electric lines of force are radial and inwards.

34. Electric lines of force

(a) exist everywhere

(b) exist only in the immediate vicinity of electric charges

(c) exist only when both positive and negative charges are near one another

(d) are imaginary

Show Answer

Answer: (d) Electric lines of force are imaginary.

35. Debye is the unit of

(a) electric flux

(b) electric dipole moment

(c) electric potential

(d) electric field intensity

Show Answer

Answer: (b) Debye is the unit of electric dipole moment.

36. An electric dipole will experience a net force when it is placed in

(a) a uniform electric field

(b) a non-uniform electric field

(c) both the above cases

(d) none of these

Show Answer

Answer: (b) Net force (as well as torque) are experienced by the dipole in a non uniform electric field.

37. An electric dipole is kept in a uniform electric field. it experiences

(a) a force and a torque

(b) a force, but no torque

(c) A torque but no force

(d) neither a force nor a torque

Show Answer

Answer: (c) In a uniform electric field, net force $=0$, but torque $\neq 0$.

38. A charge $q$ is enclosed in a cube. What is the electric flux associated with one of the faces of cube

(a) $q / \varepsilon_0$

(b) $\varepsilon_0 / q$

(c) $6 q / \varepsilon_0$

(d) $q / 6 \varepsilon_o$

Show Answer

Answer: (d). Aceording to Gauss theorem-the total electric flux $$ \phi=\frac{1}{\varepsilon_o} \times \text { total enclosed charge }=\frac{1}{\varepsilon_o} \times q . $$

Since cube has six faces. Hence electric flux linked with each face $=(1 / 6 \phi)=q / 6 \varepsilon_0$.

39. A charge $Q$ is placed at the corner of a cube. The electric flux through all the faces of the cube is

(a) $Q / \varepsilon_0$

(b) $Q / 6 \varepsilon_o$

(c) $Q / 8 \varepsilon_o$

(d) $Q / 3 \varepsilon_o$

Show Answer

Answer: (c) We require eight cubes to surround completely this corner charge. So, the flux through one cube is $$ \frac{1}{8} \times \frac{Q}{\varepsilon_o}=\frac{Q}{8 \varepsilon_o} $$

40. Positive electric flux indicates that electric lines of force are directed

(a) outwards

(b) inwards

(c) outwards or inwards

(d) none of these

Show Answer

Answer: (a) Positive electric flux is due to + charge. So lines of force are directed outwards.

41. Electric flux at a point in an electric field is

(a) positive

(b) negative

(c) zero

(d) none of these

Show Answer

Answer: (c) As area of a point is zero, $$ \therefore \phi=\mathrm{E} \text { (ds) } \cos \theta=\mathrm{E} \cos \theta \times 0=\text { Zero. } $$

42. Electric flux over a surface in an electric field may be

(a) positive

(b) negative

(c) zero

(d) positive, negative, zero

Show Answer

Answer: (d) Though electric flux is scalar, yet its value may be positive, negative or zero.

43. The electric potential at a point on the equatorial line of an electric dipole is

(a) directly proportional to distance

(b) inversely proportional to distance

(c) inversely proportional to square of the distance

(d) none of the above

Show Answer

Answer: (d) At a point on equatorial line of electric dipole, V=0.

More Than One Correct:

DIRECTIONS: This section contains multiple choice questions. Each question has 4 choices $(a),(b),(c)$ and $(d)$ out of which ONE OR MORE may be correct.

1. Four electric charges are arranged as shown in the figure at the four corners of a square of side a. The potential energy of the system is:

(a) zero

(b) negative

(c) positive

(d) greater than $\frac{1}{4 \pi \varepsilon_0} \frac{2 q^{2}}{a}$

(e) less than $\frac{1}{4 \pi \varepsilon_0}(-\frac{2 q^{2}}{a})$

Show Answer

Answer: (b,c)

2. Three charged particles are in equilibrium under their electrostatic forces only:

(a) the particles must be collinear

(b) all the charges cannot have the same magnitude

(c) all the charges cannot have the same sign

(d) the equilibrium is unstable

Show Answer

Answer: (a,b,c,d)

3. Inside a uniformly charged spherical conductor, the electric:

(a) potential is zero everywhere

(b) potential is non-zero and same everywhere

(c) field is zero everywhere

(d) field has the same magnitude everywhere but it is not zero

Show Answer

Answer: (b,c)

4. Four charges, all of the same magnitude are placed at the four corners of a square. At centre of the square, the potential is $V$ and the field is $E$. By suitable choices of the signs of four charges, which of the following can be obtained?

(a) $V=0, E=0$

(b) $V=0, E \neq 0$

(c) $V \neq 0, E=0$

(d) $V \neq 0, E \neq 0$

Show Answer

Answer: (a,b,c,d)

5. Consider two identical charges placed at a distance $2 d$ apart as shown in the figure. The equilibrium of a positive test charge placed at the point $O$ midway between them may be:

(a) neutral

(b) stable for displacement along the $x$-axis

(c) stable for displacement along the $y$-axis

(d) unstable for displacement along the $x$-axis

Show Answer

Answer: (b,c,d)

6. Across the surface of a charged conductor, the electric:

(a) field is continuous

(b) potential is continuous

(c) field is discontinuous

(d) potential is discontinuous

Show Answer

Answer: (b,c)

7. A point charge is brought in an electric field. The electric field at a nearby point:

(a) will increase if the charge is positive

(b) will decrease if the charge is negative

(c) may increase if the charge is positive

(d) may decrease if the charge is negative

Show Answer

Answer: (c,d)

8. Which of the following quantities do not depend on the choice of zero potential or zero potential energy?

(a) Potential at a point

(b) Potential difference between the points

(c) Potential energy of a two charge system

(d) Change in potential energy of a two charge system

Show Answer

Answer: (b,d)

9. An electric dipole is placed at the centre of a sphere. Mark the correct option:

(a) the flux of the electric field through the sphere is zero

(b) the electric field is zero at every point of the sphere

(c) the electric field is not zero anywhere on the sphere

(d) the electric field is zero on a circle on the sphere

Show Answer

Answer: (a,c)

10. In a uniform electric field, equipotential surfaces must:

(a) be plane surfaces

(b) be normal to the direction of the field

(c) be placed such that surfaces having equal differences in potential are separated by equal distances

(d) have decreasing potentials in the direction of the field

Show Answer

Answer: (a,b,c,d)

11. $S_1$ and $S_2$ are two equipotential surfaces on which the potentials are not equal:

(a) $S_1$ and $S_2$ both cannot intersect

(b) $S_1$ and $S_2$ both cannot be plane surfaces

(c) in the region between $S_1$ and $S_2$, the field is maximum where they are closest to each other

(d) a line of force from $S_1$ to $S_2$ must be perpendicular to both

Show Answer

Answer: (a,c,d)

12. Which of the following is/are proportional to the inverse square of the distance $x$ ?

(a) The potential at a distance $x$ from an isolated point charge

(b) The electric field at a distance $x$ from an isolated point charge

(c) The force per unit length between two thin, straight, infinitely long current carrying conductors, parallel to each other, separated by a distance $x$

(d) The gravitational attraction between two small bodies kept at a distance $x$ apart

Show Answer

Answer: (a,c)

13. A positive charge $Q$ is placed at the centre of a thin metallic spherical shell. Select the correct statements from the following:

(a) the electric field at any point outside the shell is zero

(b) the electrostatic potential at any point outside the shell is $Q / 4 \pi \varepsilon_1 r$, where $r$ is the distance of the point from $O$

(c) the outer surface of the spherical shell is an equipotential shell

(d) the electric field at any point inside the shell, other than $O$. is zero

Show Answer

Answer: (b,c)

14. A parallel plate capacitor is charged to a definite potential and the charging battery is disconnected. Now if the plates of capacitor are moved apart, then:

(a) the stored energy of the capacitor increases

(b) charge on the capacitor increases

(c) voltage of the capacitor increases

(d) the capacitance increases

Show Answer

Answer: (a,c)

15. A capacitor of capacity $100 \mu F$ is discharged through 3 resistance of $1000 ohm$ :

(a) its time constant is $1 \times 10^{-1} s$

(b) during time $0.693 \times 10^{-1} s$, the charge left on the capacitor is half of its maximum change

(c) during its time constant charge left on the capacitor is 0.63 times to its maximum charge

(d) during its time constant, charge left on the capacitor is 0.37 times to its maximum change

Show Answer

Answer: (a,b,d)

16. Two parallel plate capacitors are constructed one by a pair of iron plates and the second by a pair of copper plates of same area and same spacings. Then:

(a) the copper plate capacitor has a greater capacitance than the iron one

(b) both capacitors have equal, non-zero capacitances in the uncharged state

(c) both capacitors will have equal capacitances only if they are charged equally

(d) the capacitances of the two capacitors are equal even if they are unequally charged

Show Answer

Answer: (b,d)

17. Mark the correct statements?

(a) A given conducting sphere can be charged to any extent.

(b) A given conducting sphere cannot be changed to a potential greater than a certain valuc.

(c) A given conducting sphere cannot be charyed to a potential less than a certain minimum value.

(d) None of the above

Show Answer

Answer: (b,c)

Fill In the Passage

DIRECTIONS : Fill in the blanks in the following passage(s) from the words given inside the box.

I.

$\begin{aligned} &\text{ Friction} &\text{ electrons } &\text{ transferring } &\text{ Negative charge } \\ &\text{ Positive charge }\end{aligned}$

Electric charge is developed due to actual transfer of ……1……. When two substances are rubbed against each other, energy is provided from outside to overcome ……2……. between them. This energy is used to remove electrons from one substance and …….3……. them to the other. The transfer takes place from the material in which electrons are held less tightly to the material in which electrons are held more tightly. The material which loses electrons acquires …….4……. and which gains electrons acquires an equal ……5……..

Show Answer

Answer:

1. electrons

2. friction

3. transferring

4. positive charge

5. Negative charge

II.

$\begin{aligned} &\text{multiple} &\text{ + ne} &\text{ integral number} &\text{ electron} \\ &\text{quantization} &\text{ $\pm e$ }\end{aligned} $

Any charge exists in discrete lumps or packets of a certain minimum charge …………… where $e$ is the charge of an electron. According to …………….. of charge, the charge on a body can be only an integral …………… of . charge on the ……. 4……….., $q=\ldots . .5$…….. where $n=1,2$, $3 \ldots \ldots$. and $e=1.6 \times 10^{-19} C$ The cause of quantization is that only …………. of electrons can be transferred from a body to the other on rubbing.

Show Answer

Answer:

1. $\pm \mathrm{e}$

2. quantization

3. multiple

4. electron

5. +ne

6. integral number

III.

$\begin{aligned} &\text{ $q_1 q_2$} &\text{ Permittivity} &\text{ directly } &\text{ $r^2$} &\text{ inversely} \end{aligned} $

The force of interaction between any two point charges is …..1……… proportional to the product of the charges and …….2……. proportional to the square of the distance between them.

$ \begin{aligned} & \begin{matrix} \text{ Mathematically, } & F \propto \ldots . \\ \text{ and } & F \propto \frac{1}{r^{2}} \end{matrix} \\ & \therefore \quad F \propto \frac{q_1 q_2}{\ldots \ldots . .4 \ldots \ldots} \\ & \therefore \quad F=k \frac{q_1 q_2}{r^{2}} \end{aligned} $

where, $k=$ electrostatic force constant

$=\frac{1}{4 \pi \varepsilon_0}$ in free space $=\frac{1}{4 \pi \varepsilon}$ in any medium

where, $\varepsilon$ is the absolute electric ………….. of the medium and $\varepsilon_0$ is the absolute electric permittivity of free space or vacuum.

$\therefore \quad F=\frac{1}{4 \pi \varepsilon_0} \frac{q_1 q_2}{r^{2}}$ and $F_m=\frac{1}{4 \pi \varepsilon} \frac{q_1 q_2}{r^{2}}$ are forces

between two charges $q_1$ and $q_2$ separated by a distance ’ $r$ ’ in free space and a medium respectively.

Dividing we get, $\frac{F}{F_m}=\frac{\varepsilon}{\varepsilon_0}=\varepsilon_r$ or $k$; where $\varepsilon_r$ is called relative permittivity or $k$ is called dielectric constant or specific inductive capacity.

Show Answer

Answer:

1. directly

2. inversely

3. $q_1 q_2$

4. $r^2$

5. Permittivity

Passage Based Questions:

DIRECTIONS : Study the given paragraph(s) and answer the following questions.

Passage - I

Three large plates A, B and C are placed parallel to each other and charges are given as shown.

1. The charge that appears on the left surface of plate $B$ is:

(a) $5 C$

(b) $6 C$

(c) $3 C$

(d) $-3 C$

Show Answer

Answer: (b)

2. The charge on inner surface of plate $C$, if plate $B$ is earthed,

(a) $5 C$

(b) $6 C$

(c) $3 C$

(d) $-3 C$

Show Answer

Answer: (a)

3. The charge on left surface of $B$, if $B$ and $C$ both are earthed,

(a) $5 C$

(b) $6 C$

(c) $3 C$

(d) $-3 C$

Show Answer

Answer: (c)

Passage -II

Related to the following diagram of two charges, $+Q$ and $-4 Q$

1. The net electric field is zero near which point?

(a) $A$

(b) $B$

(c) $C$

(d) $D$

Show Answer

Answer: (a)

2. At which point does the net electric field vector point to the left?

(a) $A$

(b) $B$

(c) $C$

(d) $D$

Show Answer

Answer: (d)

3. At which point would a small positive charge $q$ feel the greatest force?

(a) $A$

(b) $B$

(c). C

(d) $D$

Show Answer

Answer: (c)

Passage-III

Two fixed charges $-2 Q$ and $Q$ are located at the points with coordinates $(-3 a, 0)$ and $(+3 a, 0)$ respectively in the $X-Y$ plane.

1. All the points in the $X-Y$ plane where the electric potential due to the two charges is zero, lie on a;

(a) straight line

(b) circle

(c) parabola

(d) ellipse

Show Answer

Answer: (b)

2. The expression for the potential $V(x)$ at a general point on the $x$-axis is given by:

(a) $V(x)=\frac{1}{4 \pi \varepsilon_0}[\frac{Q}{3 a-x}-\frac{2 Q}{(3 a+x)}]$ for $0<x<3 a$

(b) $V(x)=\frac{1}{4 \pi \varepsilon_0}[\frac{2 Q}{3 a-x}-\frac{Q}{(3 a+x)}]$ for $0<x<3 a$

(c) $V(x)=\frac{1}{4 \pi \varepsilon_0}[\frac{Q}{3 a+x}-\frac{2 Q}{(3 a-x)}]$ for $0<x<3 a$

(d) $V(x)=\frac{1}{4 \pi \varepsilon_0}[\frac{Q}{3 a-x}+\frac{2 Q}{(3 a+x)}]$ for $0<x<3 a$

Show Answer

Answer: (a)

Assertion & Reason:

DIRECTIONS : Each of these questions contains an Assertion followed by Reason. Read them carefully and answer the question on the basis of following options. You have to select the one that best describes the two statements.

(a) If both Assertion and Reason are correct and Reason is the correct explanation of Assertion.

(b) If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion.

(c) If Assertion is correct but Reason is incorrect.

(d) If Assertion is incorrect but Reason is correct.

1. Assertion : Consider two identical charges placed distance

$2 d$ apart, along $x$-axis.

The equilibrium of a positive test charge placed at the point $O$ midway between them is stable for displacements along the $x$-axis.

Reason: Force on test charge is zero.

Show Answer

Answer: (c) If +ve charge is displaced along $x$-axis, then net force will always act in a direction opposite to that of displacement and the test charge will always come back to its original position.

2. Assertion : A deuteron and an $\alpha$-particle are placed in an electric field. If $F_1$ and $F_2$ be the forces acting on them and $a_1$ and $a_2$ be their accelerations respectively then, $a_1=a_2$.

Reason: Forces will be same in electric field.

Show Answer

Answer: (d) $q_d=e, m_d=2 m_p=2 m$

$q _{\alpha}=2 e, m _{\alpha}=4 m_p=4 m$

$F_1=F _{\alpha}=eE, F_2=F _{\alpha}=2 eE \neq F_1$

Further, $\quad a_1=\frac{F_1}{2 m}=\frac{e E}{2 m}$

and $a_2=\frac{F_2}{2 m}=\frac{2 e E}{4 m}=\frac{e E}{2 m}=a_1$

3. Assertion : Four point charges $q_1, q_2, q_3$ and $q_4$ are as shown in figure. The flux over the shown Gaussian surface depends only on charges $q_1$ and $q_2$.

Reason: Electric field at all points on Gaussian surface depends only on charges $q_1$ and $q_2$.

Show Answer

Answer: (d) Electric field at any point depends on presence of all charges.

4. Assertion : The positive charge particle is placed in front of a spherical uncharged conductor. The number of lines of forces terminating on the sphere will be more than those emerging from it.

Reason: The surface charge density at a point on the sphere nearest to the point charge will be negative and maximum in magnitude compared to other points on the sphere.

Show Answer

Answer: (a) No. of lines entering the surface $=$ No. of lines leaving the surface.

5. Assertion: Charges are given to plates of two plane parallel plate capacitors $C_1$ and $C_2$ (such that $C_2=2 C_1$ ) as shown in figure. Then the key $K$ is pressed to complete the circuit. Finally the net charge on upper plate and net charge the circuit. Finally the net charge on upper plate and net charge on lower plate of capacitor $C_1$ is positive.

Reason: In a parallel plate capacitor both plates always carry equal and opposite charge.

Show Answer

Answer: (d) Charge distribution on each surface makes both capacitor of same potential difference hence charge will not flow.

6. Assertion : Each of the plates of a parallel-plate capacitor is given equal positive charge $\mathbf{Q}$. The charges on the facing surfaces will be same.

Reason: A negative charge ( $-Q$ ) will be induced on each of the facing surfaces.

Show Answer

Answer: (d) The charge on each of two facing surfaces will be zero.

7. Assertion : Electric potential and electric potential energy are different quantities.

Reason: For a system of positive test charge and point charge electric potential energy $=$ electric potential.

Show Answer

Answer: (d) Potential and potential energy are different quantities and cannot be equated.

8. Assertion : Two equipotential surfaces cannot cut each other.

Reason: Two equipotential surfaces are parallel to each other.

Show Answer

Answer: (d) Two equipotential surfaces are not necessarily parallel to each other

Multiple Matching Questions:

DIRECTIONS : Each question has four statements $(A, B, C$ and D) given in Column I and five statements ( $p, q, r, s$ and $t$ ) in Column II. Any given statement in Column I can have correct matching with one or more statement(s) given in Column II. Match the entries in column I with entries in column II.

1. For the situation shown in the figure below, match the entries of column I with the entries of column II.

$ \begin{array}{ll} \text{Column I} & \text{Column II} \\ \text{(A) If we displace the inside charge} & \text{(p) distribution of charge on the inner surface of . conductor is uniform} \\ \text{(B) If we displace the outside charge} & \text{(q) distribution of charge on the inner surface of conductor is nonuniform} \\ \text{(C) In ths situation shown} & \text{(r) distribution of charge on the outer surface of conductor is uniform} \\ \text{(D) If outside charge is not present} & \text{(s) distribution of charge on the outer surface of conductor is nonuniform} \\ \end{array} $

Show Answer

Answer: (A) $\to q, \mathrm{s;}(B) \to p, \mathrm{s;}$ (C) $\to p, \mathrm{s;}$ (D) $\to p, r$

2. In the circuit shown in figure, plates of capacitor $C_1$ are gradually pulled apart. Then, match the entries of column I with the entries of column II.

$ \begin{array}{ll} \text{Column I} & \text{Column II} \\ \text{(A) Potential difference across $C_2$.} & \text{(p) will increase} \\ \text{B) Charge on $C_2$} & \text{(q) will decrease stored in $C_2$} \\ \text{(C) Potential energy} & \text{(r) will remain same} \\ \text{(D) Electric field between plates of $C_1$} & \\ \end{array} $

Show Answer

Answer: (A) $\to q:(B) \to q ;(C) \to q:(D) \to q$

3. In the circuit shown in the following figure, $C_1=C, C_2=2 C$, $C_3=3 C, C_4=4 C$. Match the entries of column I with the entries of column II.

$ \begin{array}{ll} \text{Column I} & \text{Column II} \\ \text{(A) Maximum potential difference} & \text{(p) across $C_1$} \\ \text{(B) Minimum potential difference} & \text{(q) across $C_2$} \\ \text{(C) Maximum potential energy} & \text{(r) across $C_3$} \\ \text{(D) Minimum potential energy} & \text{(s) across $C_4$} \\ \end{array} $

Show Answer

Answer: (A) $\to p ;(B) \to r, s ;(C) \to p ;(D) \to r$

4. In the given figure, the separation between the plates of $C_1$ is slowly increased to double of its initial value, then match the following.

$ \begin{array}{ll} \text{Column I} & \text{Column II} \\ \text{(A) The potential difference across $C_1$} & \text{(p) increases} \\ \text{(B) The potential difference across $C_2$} & \text{(q) decreases} \\ \text{(C) The energy stored in $C_1$} & \text{(r) increases by a factor of $6 / 5$} \\ \text{(D) The energy stored in $C_2$} & \text{(s) decreases by a factor of $18 / 25$} \\ \end{array} $

Show Answer

Answer: (A) $\to p, r ;(B) \to q ;(C) \to q, s$; (D) $\to q$

5. Match the entries of colun I with the entries of column II. Column I Column II

Column I	Column II
(A) When a dielectric slab is gradually inserted between the plates of an isolated parallel plate capacitor	(p) is gradually inserted between the plates of an isolated parallel plate capacitor
(B) When a dielectric slab is gradully inserted between the plates of a parallel plate capacitor and its potential is kept constant	(q) is gradully inserted between the plates of a parallel plate capacitor and its potential is kept constant
(C) When the plates of a parallel plate capacitor are pulled apart, keeping its potential constant	(r) parallel plate capacitor are pulled apart, keeping its potential constant
(D) When the plates of a parallel plate capacitor are pulled apart, keeping its charge constant.	(s) work done by extrnal agent is negative

Show Answer

Answer: (A) $\to p, s ;(B) \to r, s ;(C) \to p, q ;(D) \to q$

Subjective Questions:

DIRECTIONS : Answer the following questions.

1. If a point charge $+q$, is taken first from $A$ to $C$, then from $C$ to $B$ of a circle drawn with another point charge $+q$ at centre, then along which path more work will be done?

Show Answer

Answer: Work done will be same in both the paths as electrostatic force is conservative in nature, so work done doesn’t depend on the path followed.

2. Find the electric field between two metal plates $3 mm$ apart, connected to a $12 V$ battery.

Show Answer

Answer: $E=\frac{d V}{d r}=\frac{12}{3 \times 10^{-3}}=4 \times 10^{3} N / C$.

3. Calculate the electric potential at the surface of a gold nucleus. Given, the radius of the nucleus $=6.6 \times 10^{-15} m$ and atomic number of gold $=79$.

Show Answer

Answer: Charge on gold nucleus $=q=Ze=79 \times 1.6 \times 10^{-19}$

$ V=\frac{kq}{r}=\frac{9 \times 10^{9} \times 79 \times 1.6 \times 10^{-19}}{6.6 \times 10^{-15}}=1.724 \times 10^{7} V $

4. Calculate the potential at a point $P$ due to a charge of $4 \times 10^{-7} C$ located $10 cm$ away.

Show Answer

Answer: $\quad V=\frac{kq}{r}=\frac{9 \times 10^{9} \times 4 \times 10^{-7}}{0 . r}=3.6 \times 10^{4} V$.

5. At a point due to a point charge, the values of electric field intensity and potential are $32 N / C$ and $16 J / C$ respectively. Calculate magnitude of charges and distance of charge from the point of observation.

Show Answer

Answer: $E=\frac{kq}{r^{2}}=32$ and $V=\frac{kq}{r}=16$

$ r=\frac{16}{32}=0.5 m \therefore \frac{kq}{0.5}=16 \Rightarrow q=8.89 \times 10^{-10} C $

6. A parallel plate capacitor with air has a capacitance of $10 \mu F$. If the distance between the plates is reduced to half and the space between them is filled with a material of dielectric constant 10 , find the new capacitance.

Show Answer

Answer: $C_1=\frac{\varepsilon_0 A}{d}=10 p F, C_2=\frac{k \varepsilon_0 A}{d / 2}$

$ =10 \times 10 \times 2 \mu F=200 \mu F $

7. Each plate of a parallel plate capacitor has an area of $6 \times 10^{-3} m^{2}$ and the distance between the plates is $3 mm$. Calculate the capacitance of the capacitor with air as dielectric. If this capacitor is connected to a $200 V$ supply, what is the change on each plate of the capacitor?

Show Answer

Answer: $C=\frac{\varepsilon_0 A}{d}=\frac{8.85 \times 10^{-12} \times 6 \times 10^{-3}}{3 \times 10^{-3}}=17.7 \mu F$

$ q=CV=17.7 \times 10^{-12} \times 200=3.54 \times 10^{-9} C $

8. A given charge situated at a distance from an electric dipole in the end on position, experiences a force $F$. If the distance of charge is doubled, what will the force acting on the charge?

Show Answer

Answer: For an electric dipole, $F \infty \frac{1}{r^{3}}$

$\therefore \quad$ New force, $F^{1}=\frac{F}{2^{3}}=\frac{F}{8}$.

9. In a parallel plate capacitor the potential difference of $10^{2} V$ is maintained between the plates. What will be the electric field at points $A$ and $B$ ?

Show Answer

Answer:

Electric field between the plates of a capacitor is uniform that is constant at every point $(E=\frac{\sigma}{\varepsilon_0})$.

$\therefore \quad$ Electric field is same at $A$ and $B$.

10. Two protons $A$ and $B$ are placed between two parallel plates having a potential difference of $V$ as shown in the figure. Will these proton experience equal or unequal force?

Show Answer

Answer:

The field between the two parallel plates is uniform. So, the protons will experience same force.

11. $A$ and $B$ are two conducting spheres of the same radius. $A$ being solid and $B$ hollow, both are charged to the same potential. What will be the relation between the charges on the two spheres?

Show Answer

Answer: Charge will be same, as $Q=CV=4 \pi \varepsilon_0 RV$.

Here, $R$ and $V$ both are same.

12. How much energy will be stored by a capacitor of $470 \mu F$ when charged by a battery of $20 V$ ?

Show Answer

Answer: Energy $=\frac{1}{2} CV^{2}=\frac{1}{2} \times 470 \times 10^{-6} \times(20)^{2}=94 \times 10^{-3} J$.

13. How does the force between two point charges change if the dielectric constant of the medium in which they are kept increases?

Show Answer

Answer: Force in a medium $=F_m=\frac{F}{k}$

14. Is electrostatic potential a scalar quantity or a vector quantity?

Show Answer

Answer: It is a scalar quantity.