(a) Describe briefly the process of transferring the charge between the two plates of a parallel plate capacitor when connected to a battery. Derive an expression for the energy stored in a capacitor.

(b) A parallel plate capacitor is charged by a battery to a potential difference V. It is disconnected from battery and then connected to another uncharged capacitor of the same capacitance. Calculate the ratio of the energy stored in the combination to the initial energy on the single capacitor.
OR
(a) Derive an expression for the electric field at any point on the equatorial line of an electric dipole.

(b) The identical point charges, q each, are kept 2 m apart in air. A third point charge Q of unknown magnitude and sign is placed on the line joining the charges such that the system remains in equilibrium. Find the position and nature of Q.
 

(a) When the plates of the parallel plate capacitor is connected to a battery. Then the first insulated metal plate gets, the positive charge till its potential become maximum. Then, the charge will leak to surroundings. So, the negative charge will be induced on the nearer face of the second plate and the positive charge will be induced on its farther plate.

Consider a capacitor of capacitance C. Initial charge and potential difference be zero. Let, a charge Q be given in small steps. Let at any instant when charge on capacitor be q, the potential difference between its plates,
$\mathrm{P}=\frac{\mathrm{q}}{\mathrm{C}}$
Now work done in giving an additional charge dg is,
$\mathrm{dW}=\mathrm{Vdq}=\frac{\mathrm{q}}{\mathrm{C}}\mathrm{dq}$
Total work done in giving charge from 0 to Q is 
$\begin{array}{l}\mathrm{W}={\int }_0^{\mathrm{Q}} \mathrm{Vdq}={\int }_0^{\mathrm{Q}} \frac{\mathrm{q}}{\mathrm{C}}\mathrm{dq}\\ =\frac{1}{\mathrm{C}}{\left[\frac{{\mathrm{q}}^2}{2}\right]}_0^{\mathrm{Q}}=\frac{1}{\mathrm{C}}\left[\frac{{\mathrm{Q}}^2}{2}-\frac{0}{2}\right]=\frac{{\mathrm{Q}}^2}{2\mathrm{C}}\\ \mathrm{W}=\frac{(\mathrm{CV}{)}^2}{2\mathrm{C}}=\frac{1}{2}{\mathrm{CV}}^2=\frac{1}{2}\mathrm{QV}[\because \mathrm{Q}=\mathrm{CV}]\end{array}$
$\therefore$Electrostatic potential energy,
$$\begin{gathered} \left(\mathrm{b}\right) \mathrm{Here}, {\mathrm{C}}_1=\mathrm{C},{\mathrm{C}}_2=\mathrm{C} \\ {\mathrm{V}}_1=\mathrm{V},{\mathrm{V}}_2=0 \end{gathered}$$
Now initially, energy stored in first capacitor, as second capacitor is uncharged.
$\begin{array}{l}{\mathrm{U}}_1=\frac{1}{2}{\mathrm{CV}}_1^2\\ =\frac{1}{2}{\mathrm{CV}}^2 .....\left(\mathrm{i}\right)\end{array}$
OR
(a) Derive an expression for the electric field at any point on the equatorial line of an electric dipole.
Consider an electric dipole of charges -q and +q separated by a distance 2a and placed in a free space. Let P be a point on equitorial line of dipole at a distance r from the centre of a dipole.

Let, E_A and E_B be the electric field at point
P due to charges -q and +q
Then resultant electric field at point P is
$\overrightarrow{\mathrm{E}}={\overrightarrow{\mathrm{E}}}_{\mathrm{A}}+{\overrightarrow{\mathrm{E}}}_{\mathrm{B}}$
Now,
$\begin{array}{r}\left|{\overrightarrow{\mathrm{E}}}_{\mathrm{A}}\right|=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\cdot \frac{\mathrm{q}}{{\mathrm{AP}}^2}=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{\mathrm{q}}{\left({\mathrm{r}}^2+{\mathrm{a}}^2\right)}\text{ (along PA) }\\ \left|{\overrightarrow{\mathrm{E}}}_{\mathrm{B}}\right|=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\cdot \frac{\mathrm{q}}{{\mathrm{BP}}^2}=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{\mathrm{q}}{\left({\mathrm{r}}^2+{\mathrm{a}}^2\right)}\text{ (along BP) }\end{array}$
The, resultant intensity is the vector sum of E_A and E_B
E_A and E_B can be resolved into two components.
The Y-components cancel out each other. And X-component will add up to give the resultant field.

$\therefore \left|\overrightarrow{\mathrm{E}}\right|={\mathrm{E}}_{\mathrm{A}}\cos \mathrm{\theta }+{\mathrm{E}}_{\mathrm{B}}\cos \mathrm{\theta }$
Now in right, triangle ORB
$\begin{array}{l}\cos \mathrm{\theta }=\frac{\mathrm{OB}}{\mathrm{BP}}=\frac{\mathrm{a}}{\sqrt{{\mathrm{r}}^2+{\mathrm{a}}^2}}\\ \therefore \mathrm{E}=2\times \frac{1}{4{\mathrm{\pi \epsilon }}_0}\times \frac{\mathrm{q}}{{\mathrm{r}}^2+{\mathrm{a}}^2}\frac{\mathrm{a}}{{\left({\mathrm{r}}^2+{\mathrm{a}}^2\right)}^{1/2}}\end{array}$
$\begin{array}{l}=\frac{2\mathrm{qa}}{4{\mathrm{\pi \epsilon }}_0{\left({\mathrm{r}}^2+{\mathrm{a}}^2\right)}^{3/2}}\\ =\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{\mathrm{p}}{{\left({\mathrm{r}}^2+{\mathrm{a}}^2\right)}^{3/2}}\\ [\because 2\mathrm{qa}=\mathrm{p}]\end{array}$
This is the required expression.
(b) Let the two charges of + q each placed at point A and B at a distance 2 m apart in air.


Suppose, the third charge Q (unknown magnitude and charge) is placed at a point O, on the line joining the other two charges, such that OA= x and OB 2-x. 

For the system to be in equilibrium, net force on each 3 charges must be zero.

If we assume that charge Q placed at O is positive, the force on it at O may be zero. But the force on charge q at point A or B will not be zero. It is because, the forces on a charge q due to the other two charges will act in same direction. 
If charge Q is negative, then the forces on q due to other two charges will act in opposite direction.
Hence, Q will be negative in nature. For charge (-Q) to be in equilibrium Force on charge (-q) due to charge (+q) at point A should be equal and opposite to charge (+Q) at B

$\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{\mathrm{Qq}}{{\mathrm{x}}^2}=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\cdot \frac{\mathrm{Qq}}{(2-\mathrm{x}{)}^2}$
or    $(2-\mathrm{x}{)}^2={\mathrm{x}}^2$
$\Rightarrow \mathrm{x}=(2-\mathrm{x})\Rightarrow \mathrm{x}=1\mathrm{m}$
Therefore, for the system to be in equilibrium a charge – Q is placed at a mid point between the two charges of + q each.