Solution:

Ans. (i) Finding the electrostatic potential

(ii) Finding the work done

(iii) Effect of change of path

(iv) Potential energy of the system

1 (with justification in each case)

(i) We have, for a point charge,

$$ V=\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{q}{r} $$

(a) At point $(0,0, z)$ :

Potential due to the charge $(+q)$

$$ V_{+}=\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{q}{(z+a)} $$

Potential due to the charge $(-q)$

$$ V_{-}=\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{(-q)}{(z-a)} $$

Total Potential at $(0,0, z)$

$$ \begin{align*} & =\frac{1}{4 \pi \varepsilon_{0}}\left[\frac{1}{z+a}-\frac{1}{z-a}\right] \ & =\frac{-2 q a}{4 \pi \varepsilon_{0}\left(z^{2}-a^{2}\right)} \end{align*} $$

(b) At point $(x, y, 0)$

Potential due to the charge $+q$

$$ V_{+}=\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{q}{\sqrt{x^{2}+y^{2}+a^{2}}} $$

Potential due to the charge $(-q)$

$$ V_{-}=\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{-q}{\sqrt{x^{2}+y^{2}+a^{2}}} $$

Total potential at $(x, y, 0)$

$=\frac{q}{4 \pi \varepsilon_{0}}\left(\frac{1}{\sqrt{x^{2}+y^{2}+a^{2}}}-\frac{1}{\sqrt{x^{2}+y^{2}+a^{2}}}\right)=0$ Note : Give full credit of part (b) if a student writes that the point $(x, y, 0)$ is equidistant from charges $+q$ and $-q$, Hence total potential due to them at the given point will be zero.

(ii) Work done $=q\left[V_{1}-V_{2}\right]$

$$ V_{1}=0 \text { and } V_{2}=0 $$

$\therefore \quad$ Work done $=0$

Where $V_{1}$ and $V_{2}$ are the total potential due to dipole at point $(5,0,0)$ and $(-7,0,0)$

$1 / 2$

(iii) There would be no change

This is because the electrostatic field is a conservative field.

1

(Alternatively : The work done, in moving a test charge between two given points is independent of the path taken)

(iv) The two given charges make an electric dipole of dipole moment $p=q \cdot \overrightarrow{2 a}$

P.E. in position of unstable equilibrium (where $\vec{p}$ and $\vec{E}$ antiparallel to each other) $1 / 2$

$=P \mathrm{p}=2 a q E$

$1 / 2$

[CBSE Marking Scheme, 2018]

Revision Notes

Conductors and insulators

Conductors are the materials through which charge can move freely. Examples : metals, semi-metals as carbon, graphite, antimony and arsenic.

• Insulators are materials in which the electrical current will not flow easily. Such materials cannot be grounded and do not easily transfer electrons. Examples : plastics and glass.

Dielectrics

TOPIC-2

Capacitance

• These are the material in which induced dipole moment is linearly proportional to applied electric field.
• Electrical displacement or electrical flux density $D=\varepsilon_{r} \varepsilon_{0} E$.

where, $\varepsilon_{\mathrm{r}}=$ relative permittivity, $\varepsilon_{0}=$ permittivity of free space and $E$ is electric field.

$>$ If a dielectric is kept in between the plates of capacitor, capacitance increases by factor ’ $\kappa$ ’ (kappa) known as dielectric constant, so $C=\kappa \varepsilon_{0} \frac{A}{d}$

where, $\mathrm{A}=$ area

$\kappa=$ dielectric constant of material also called relative permittivity $\kappa=\varepsilon_{r}=\frac{\varepsilon}{\varepsilon_{0}}$

In dielectric, polarisation and production of induced charge takes place when dielectric is kept in an external electric field.

Electric polarization

• Electric polarization $P$ is the difference between electric fields $D$ (induced) and $E$ (imposed) in dielectric due to bound and free charges written as $P=\frac{D-E}{4 \pi}$

In term of electric susceptibility : $P=\chi_{e} E$

In MKS : $P=\varepsilon_{0} \chi_{e} E$,

The dielectric constant $\kappa$ is always greater than 1 as $\chi_{e}>0$

Capacitor

• A capacitor is a device which is used to store charge.
• Amount of charge ’ $Q$ ’ stored by the capacitor depends on voltage applied and size of capacitor.
• Capacitor consists of two similar conducting plates placed in front of each other where one plate is connected to positive terminal while other plate is connected to negative terminal.
• Electric charge stored between plates of capacitor is directly proportional to potential difference between its plates, i.e.,

$$ Q=C V $$

where, $C=$ Capacitance of capacitor, $\mathrm{V}=$ potential difference between the plates

In capacitor, energy is stored in the form of electrical energy, in the space between the plates.

Capacitance

• Capacitance of a capacitor is ratio of magnitude of charge stored on the plate to potential difference between the plates, written as $C=\frac{Q}{\Delta V}$

where, $C=$ capacitance in farads $(F), Q=$ charge in Coulombs $(C), \Delta V=$ electric potential difference in Volts $(V)$,

$>$ SI unit of capacitance is farad $(F)$

$1 F=\frac{1 C}{1 V}=9 \times 10^{11}$ stat farad,

Where, stat-farad is electrostatic unit of capacitance in C.G.S. system

Capacitance of a conductor depends on size, shape, medium and other conductors in surrounding.

Parallel plate capacitor with dielectric among its plates has capacitance which is given as :

where, $\varepsilon_{0}=8.85 \times 10^{-12} \mathrm{~F} / \mathrm{m}$

Capacitor having capacitance 1 Farad is too large for electronics applications, so components with lesser values of capacitance such as $\mu_{s}$ (micro), $n$ (nano) and $p$ (pico) are applied such as :

Combination of capacitors in series and parallel

Capacitors in series

(i) If a number of capacitors of capacitances $C_{1}, C_{2}, C_{3}, \ldots \ldots \ldots \ldots . . . . C_{n}$ are connected in series, then their equivalent capacitance is given by :

$$ C=\kappa \varepsilon_{0} \frac{A}{d} $$

$$ \frac{1}{C_{s}}=\frac{1}{C_{1}}+\frac{1}{C_{2}}+\frac{1}{C_{3}}+\ldots \ldots \ldots \ldots \ldots \frac{1}{C_{n}} $$

In series combination, the charge on each capacitor is same, but the potential difference on each capacitor depends on their respective capacitance, i.e.,

$$ q_{n}=q $$

If $V_{1}, V_{2}, V_{3}, \ldots \ldots . . ., V_{n}$ be the potential differences across the capacitors and $V$ be the emf of the charging battery, then

$$ V=V_{1}+V_{2}+V_{3}+\ldots \ldots \ldots \ldots . .+V_{n} $$

As charge on each capacitor is same, therefore

$$ q=V_{1} C_{1}=V_{2} C_{2}=V_{3} C_{3} $$

the potential difference is inversely proportional to the capacitance, i.e.,

$$ V \propto \frac{1}{C} $$

In series, potential difference across largest capacitance is minimum.

The equivalent capacitance in series combination is less than the smallest capacitance in combination.

Capacitors in parallel

(i) If a number of capacitors of capacitances $C_{1}, C_{2}, C_{3} \ldots \ldots \ldots \ldots . . . C_{n}$ are connected in parallel, then their equivalent capacitance is given by,

$$ C_{p}=C_{1}+C_{2}+C_{3}+\ldots \ldots \ldots \ldots \ldots . .+C_{n} $$

• In parallel combination, the potential difference across each capacitor is same and equal to the emf of the charging battery, i.e.,

$$ V_{1}=V_{2}=V_{3}=\ldots \ldots \ldots \ldots \ldots .=V_{n}=V $$

while the charge on different capacitors may be different.

If $q_{1}, q_{2}, q_{3}, \ldots \ldots . . . ., q_{n}$ be the charges on the different capacitors, then

$$ q_{1}+q_{2}+q_{3}+\ldots \ldots \ldots \ldots \ldots+q_{n}=V C_{p} $$

As potential drop across each capacitor is same, so

$$ \Rightarrow \quad V=\frac{q_{1}}{C_{1}}=\frac{q_{2}}{C_{2}}=\frac{q_{3}}{C_{3}}=\frac{q_{n}}{C_{n}} $$

• The charges on capacitors are directly proportional to capacitances, i.e., $q \propto C$

Parallel combination is useful when largecapacitance with large charge gets accumulated on combination.

Force of attraction between parallel plate capacitor will be $F=\frac{1}{2}\left[\frac{Q V}{d}\right]=\frac{1}{2} Q E$ where $Q$ is charge on capacitor.

Capacitance of parallel plate capacitor with and without dielectric medium between the plates

• Parallel plate capacitor is a dapaditor with two identical plane parallel plates separated by a small distance where space between them is filled by dielectric medium
• The electric field between two large parallel plates is given as :

$$ E=\frac{\sigma}{\varepsilon}, $$

Where, $\sigma=$ charge density and $\varepsilon=$ permittivity Surface charge density,

$$ \sigma=\frac{Q}{A}, $$

where, $Q=$ charge on plate and $A=$ plate area

Capacitance of parallel-plate capacitor with area $A$ separated by a distance $d$ is written as $C=\varepsilon_{r} \varepsilon_{0} \frac{A}{d}$

If a dielectric slab is placed in between the plates of a capacitor, then its capacitance will increase by certain amount.

• Capacitance of parallel plate capacitor depends on plate area $A$, distance $d$ between the plates, medium between the plates ( $\kappa$ ) and not on charge on the plates or potential difference between the plates.
• If we have number of dielectric slabs of same area as the plates of the capacitor and thicknesses $t_{1}, t_{2}, t_{3}, \ldots$ and dielectric constant $\kappa_{1}, \kappa_{2}, \kappa_{3} \ldots .$. . between the plates, then the capacitance of the capacitor is given by

$$ C=\frac{\varepsilon_{0} A}{\frac{t_{1}}{\kappa_{1}}+\frac{t_{2}}{\kappa_{2}}+\frac{t_{3}}{\kappa_{3}}+\ldots .} $$

Where, $d=t_{1}+t_{2}+t_{3}+\ldots$.

If slab of conductor of thickness $t$ is introduced between the plates, then

$$ \begin{aligned} & C=\frac{\varepsilon_{0} A}{\frac{t}{\kappa}+\frac{(d-t)}{1}}=\frac{\varepsilon_{0} A}{\frac{t}{\infty}+\frac{(d-t)}{1}} \ & C=\frac{\varepsilon_{0} A}{d-t} \end{aligned} $$

$$ (\because \kappa=\infty \text { for a conductor }) $$

• When the medium between the plates consists of slabs of same thickness but areas $A_{1}, A_{2}, A_{3}, \ldots$ and dielectric constants $\kappa_{1}, \kappa_{2}, \kappa_{3} \ldots$, then capacitance is given by

$$ \begin{array}{rlrl} C & =\frac{\varepsilon_{0}\left(\kappa_{1} A_{1}+\kappa_{2} A_{2}+\kappa_{3} A_{3} \ldots\right)}{d} \ \therefore \quad & \kappa & =\frac{C_{m}}{C_{0}}=\frac{\text { capacitance in medium }}{\text { capacitance in vacuum }} \end{array} $$

D When space between the plates is partly filled with medium of thickness $t$ and dielectric constant $\kappa$, then capacitance will be :

$$ C=\frac{\varepsilon_{0} A}{d+t^{t}+\frac{t}{\kappa} d-t\left(1-\frac{1}{\kappa}\right)} $$

When there is no medium between the plates, then $\kappa=1$, so

$$ \text { C. } C_{\text {vacuum }}=\frac{\varepsilon_{0} A}{d} $$

Capacitance of spherical conductor radius $R$ in a medium of dielectric constant $\kappa$ is given by

$$ C=4 \pi \varepsilon_{0} \kappa R $$

Energy stored in capacitor

In capacitor, energy gets stored when a work is done on moving a positive charge from negative conductor to positive conductor against the repulsive forces.

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

Polar atom : Atom in which positive and negative charges possess asymmetric charge distribution about its centre.

Polarisation : The stretching of atoms of a dielectric slab under an applied electric field.

• Dielectric strength : The maximum value of electric field that can be applied to dielectric without its electric breakdown.

Dielectric : It is an electrically insulated or non-conducting material considered for its electric susceptibility.

Permittivity : It is a property of a dielectric medium that shows the forces which electric charges placed in medium exerts on each other.

OR

It is the measure of resistance that is encountered when forming an electric field in a particular medium. More specifically, permittivity describes the amount of charge needed to generate one unit of electric flux in a particular medium.

Key Formulae

Capacitance, $C=\frac{Q}{V}$, measured in Farad; $1 \mathrm{~F}=1$ coulomb/volt

Parallel plate capacitor :

$$ C=\kappa \varepsilon_{0} \frac{A}{d} $$

Cylindrical capacitor :

$$ C=2 \pi \kappa \varepsilon_{0} \frac{L}{\ln (b / a)} $$

where, $L=$ length $[\mathrm{m}], b=$ radius of the outer conductor $[\mathrm{m}], a=$ radius of the inner conductor $[\mathrm{m}]$

Spherical capacitor :

$$ C=4 \pi \kappa \varepsilon_{0}\left(\frac{a b}{b-a}\right) $$

where, $b=$ radius of the outer conductor $[\mathrm{m}], a=$ radius of the inner conductor $[\mathrm{m}]$

Maximum charge on a capacitor :

$$ Q=V C $$

For capacitors connected in series, the charge $Q$ is equal for each capdcitor as well as for the total equivalent. If the dielectric constant $\kappa$ is changed, the capacitance is multiphed by $\kappa$, the voltage is divided by $\kappa$ and $Q$ is unchanged. In vacuum $\kappa=1$ and when dielectrics are used, replace $\varepsilon_{0}$ with $\kappa \varepsilon_{0}$.

Electrical energy stored in a capacitor : [Joules (J)]

$$ U_{E}=\frac{Q V}{2}=\frac{C V^{2}}{2}=\frac{Q^{2}}{2 C} $$

Surface charge density or Charge per unit area $\mathrm{C} / \mathrm{m}^{2}$ ]

Energy density :

• Electric energy density is also called Electrostatic pressure.
• Electric force between plates of capacitor

$$ F=\frac{1}{2} \varepsilon_{0} E^{2} \cdot A $$

• Energy stored in terms of Energy density

$$ \begin{aligned} \frac{E}{A \times d} & =\frac{1}{2} \varepsilon_{0} E^{2} \ U & =\frac{1}{2} \varepsilon_{0} E^{2} \end{aligned} $$

where, $U=$ energy per unit volume $\left[\mathrm{J} / \mathrm{m}^{3}\right], \varepsilon_{0}=$ permittivity of free space, $=8.85 \times 10^{-12} \mathrm{C}^{2} / \mathrm{Nm}^{2}, E=$ energy $[\mathrm{J}]$

• Capacitors in series :

$$ \frac{1}{C_{e f f}}=\frac{1}{C_{1}}+\frac{1}{C_{2}} \cdots $$

- Capacitors in parallel :

$$ C_{e f f}=C_{1}+C_{2} \cdots $$

Objective Type Questions