Solution:

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

an additional charge $(dq)$, is

$$dW=V\cdot dq$$

But

$$\begin{array}{rlr}V& =\frac{q}{C}dW& =\frac{q}{C}dq\end{array}$$

$1/2$

Total work done in storing the charge $q$,

$$\begin{array}{rlr}\int dW& ={\int }_0^q\frac{q}{C}dqW& =\frac{1}{C}{\left(\frac{q^2}{2}\right)}_0^q=\frac{q^2}{2C}\end{array}$$

This work is stored as electrostatic energy in the capacitor.

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

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

$$\begin{array}{rlr}& U=\frac{\frac{1}{2}\left(\frac{{\epsilon }_{0}A}{d}\right)(Ed{)}^2}{Ad}& U=\frac{1}{2}{\epsilon }_0{E}^2\end{array}$$

(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)

$$\begin{array}{}\text{(ii)}& C& =\frac{{\epsilon }_{0}A}{d}C& =\frac{8.85\times {10}^{-12}\times 6\times {10}^{-3}}{3\times {10}^{-3}}\mathrm{F}C& =17.7\times {10}^{-12}\mathrm{F}=(17.7\mathrm{pF})Q& =CVQ& =17.7\times {10}^{-12}\times 100\mathrm{C}Q& =17.7\times {10}^{-10}\mathrm{C}=1.77\mathrm{nC}{Q}^{\prime }& =\kappa Q& =6\times 17.7\times {10}^{-10}\mathrm{C}& =106.2\times {10}^{-10}\mathrm{C}=10.62\times {10}^{-9}& =10.62\mathrm{nC}1/2\end{array}$$

[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

$$\begin{array}{llll}& E=\frac{\sigma }{{\epsilon }_0}=\frac{Q}{A{\epsilon }_0}\therefore & V=Ed=\frac{Qd}{A{\epsilon }_0}\text{ Capacitance, }& C=\frac{Q}{V}=\frac{{\epsilon }_{0}A}{d}\end{array}$$

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,

$$\begin{array}{rlrl}& \frac{{\sigma }_1}{{\sigma }_2}=\frac{q_1/4\pi R_1^2}{q_2/4\pi R_2^2}& \frac{{\sigma }_1}{{\sigma }_2}=\frac{q_1{R}_2^2}{q_2{R}_1^2}& \frac{{\sigma }_1}{{\sigma }_2}=\frac{R_1}{R_2}\times \frac{R_2^2}{R_1^2}=\frac{R_2}{R_1}\end{array}$$

[CBSE Marking Scheme, 2014]