Solution:

Ans. (i) Work done by the source of potential, in storing

an additional charge $(d q)$, is

$$ d W=V \cdot d q $$

But

$$ \begin{aligned} V & =\frac{q}{C} \ d W & =\frac{q}{C} d q \end{aligned} $$

$1 / 2$

Total work done in storing the charge $q$,

$$ \begin{align*} \int d W & =\int_{0}^{q} \frac{q}{C} d q \ W & =\frac{1}{C}\left(\frac{q^{2}}{2}\right)_{0}^{q}=\frac{q^{2}}{2 C} \end{align*} $$

This work is stored as electrostatic energy in the capacitor.

$$ \therefore \quad U=\frac{1}{2} C V^{2} \quad(\because q=C V) $$

Energy stored per unit volume $=\frac{\frac{1}{2} C V^{2}}{A d}$

$$ \begin{aligned} & U=\frac{\frac{1}{2}\left(\frac{\varepsilon_{0} A}{d}\right)(E d)^{2}}{A d} \ & U=\frac{1}{2} \varepsilon_{0} E^{2} \end{aligned} $$

(ii) Work done in moving the charge $q$ from $a$ to $b$, and from $c$ to $d$ is zero because electric field is perpendicular to the displacement.

Work done from $b$ to $c=-$ Work done from $d$ to $a$

$\therefore$ Total work done in moving a charge $q$ over the closed loop $=0$

[CBSE Marking Scheme 2014]

(AI Q. 22. (i) Derive the expression for the capacitance of a parallel plate capacitor having plate area $A$ and plate separation $d$.

(ii) Two charged spherical conductors of radii $R_{1}$ and $R_{2}$ when connected by a conducting wire acquire charges $q_{1}$ and $q_{2}$ respectively. Find the ratio of their surface charge densities in terms of their radii. A [Delhi I, II, III 2014] [NCERT Exemplar]

Ans. (i)

(iii)

Ans. (i)

$$ \begin{align*} C & =\frac{\varepsilon_{0} A}{d} \ C & =\frac{8.85 \times 10^{-12} \times 6 \times 10^{-3}}{3 \times 10^{-3}} \mathrm{~F} \ C & =17.7 \times 10^{-12} \mathrm{~F}=(17.7 \mathrm{pF}) \ Q & =C V \tag{ii}\ Q & =17.7 \times 10^{-12} \times 100 \mathrm{C} \ Q & =17.7 \times 10^{-10} \mathrm{C}=1.77 \mathrm{nC} \ Q^{\prime} & =\kappa Q \ & =6 \times 17.7 \times 10^{-10} \mathrm{C} \ & =106.2 \times 10^{-10} \mathrm{C}=10.62 \times 10^{-9} \ & =10.62 \mathrm{nC} 1 / 2 \end{align*} $$

[CBSE Marking Scheme 2014]

[A] Q. 24. A capacitor of unknown capacitance is connected across a battery of $V$ volts. The charge stored in it is $360 \mu C$. When potential across the capacitor is reduced by $120 \mathrm{~V}$, the charge stored in it becomes $120 \mu$

Calculate :

(i) The potential $V$ and the unknown capacitance $C$.

Surface

charge density $-\sigma$

Electric field between the plates of capacitor

$$ \begin{array}{ll} & E=\frac{\sigma}{\varepsilon_{0}}=\frac{Q}{A \varepsilon_{0}} \ \therefore & V=E d=\frac{Q d}{A \varepsilon_{0}} \ \text { Capacitance, } & C=\frac{Q}{V}=\frac{\varepsilon_{0} A}{d} \end{array} $$

i) When the two charged spherical conductors/are connected by a conducting wire, they acquire the same potential.

i.e.,

$$ \frac{K q_{1}}{R_{1}}=\frac{K q_{2}}{R_{2}} \Rightarrow \frac{q_{1}}{q_{2}}=\frac{R_{1}}{R_{2}} $$

Hence, ratio of surface charge densities,

$$ \begin{align*} & \frac{\sigma_{1}}{\sigma_{2}}=\frac{q_{1} / 4 \pi R_{1}^{2}}{q_{2} / 4 \pi R_{2}^{2}} \ & \frac{\sigma_{1}}{\sigma_{2}}=\frac{q_{1} R_{2}^{2}}{q_{2} R_{1}^{2}} \ & \frac{\sigma_{1}}{\sigma_{2}}=\frac{R_{1}}{R_{2}} \times \frac{R_{2}^{2}}{R_{1}^{2}}=\frac{R_{2}}{R_{1}} \end{align*} $$

[CBSE Marking Scheme, 2014]