a)  Deduce the expression for the electrostatic energy stored in a capacitor of capacitance ‘C’ and having charge ‘Q’. How will the 

(i) energy stored and 

(ii) the electric field inside the capacitor be affected when it is completely filled with a dielectric material of dielectric constant ‘K’? 

(b) A capacitor of unknown capacitance is connected across a battery of V volts. The charge stored in it is 360 μC. When potential across the capacitor is reduced by 120 V, the charge stored in it becomes 120 μC.
Calculate:

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

(ii) What will be the charge stored in the capacitor, if the voltage applied had increased by 120 V?

Potential of capacitor $=\frac{\mathrm{q}}{\mathrm{c}}$

$\mathrm{dW}=\frac{\mathrm{q}}{\mathrm{c}}\times \mathrm{dq}$

Small amount of work done in giving an additional charge dq to the capacitor, Total work done in giving a charge Q to the capacitor As electrostatic force is conservative, thus work is stored in the form of potential

$\mathrm{W}={\int }_{\mathrm{q}=0}^{\mathrm{q}=\mathrm{Q}} \frac{\mathrm{q}}{\mathrm{C}}=\frac{1}{2}{\left[\frac{{\mathrm{q}}^2}{\mathrm{C}}\right]}_{\mathrm{q}=0}^{\mathrm{q}=\mathrm{Q}}\mathrm{ }\therefore \mathrm{W}=\frac{1}{\mathrm{C}}\frac{{\mathrm{Q}}^2}{2}$

$\begin{array}{rr}\mathrm{U}& =\mathrm{W}=\frac{1}{2}\frac{{\mathrm{Q}}^2}{\mathrm{C}}\\ \text{ Put }\mathrm{Q}& =\mathrm{CV}\\ \therefore \mathrm{U}& =\frac{1}{2}{\mathrm{CV}}^2\\ \text{ (i) }\mathrm{C}& ={\mathrm{C}}_0\mathrm{K}\end{array}$

where ${\left[{\mathrm{E}}_0=\frac{1}{2}{\mathrm{C}}_0\mathrm{V}\right]}_1^{1+}$

Energy stored $=\frac{1}{2}{\mathrm{CV}}^2=\frac{1}{2}{\mathrm{KC}}_0{\mathrm{V}}^2$ 

 $\underset{\downarrow }{E}={\mathrm{KE}}_0$ 

energy stored

$\therefore$ Electric field will become $\frac{1}{\mathrm{K}}$ times its initial value.

(b) (i) Let the capacitance be C

$\therefore$Charge on ${\mathrm{Q}}_1=\mathrm{CV}$ or $360\mu \mathrm{C}=\mathrm{CV}\dots$ (i)

In second case,

$\begin{array}{rr} & {\mathrm{Q}}_2=\mathrm{C}(\mathrm{V}-120)\\ & \Rightarrow 120\mu \mathrm{C}=\mathrm{C}(\mathrm{V}-120)\dots \text{ (ii) }\end{array}$

From equation (i) and (ii)

$3=\frac{V}{(V-120)}$

$3\mathrm{V}-360=\mathrm{V}\Rightarrow 2\mathrm{V}=360$

By putting this value of V in (ii)

$$\begin{gathered} 120\times {10}^{-6}=C(180-120) \\ C = \frac{ 120\times {10}^{-6}}{60} = 2\mu F \end{gathered}$$

(ii) Charge stored when voltage is increased by $120\mathrm{V}$

${\mathrm{Q}}^{\mathrm{\text{'}}}=2\mu \mathrm{F}\times (180+120)\mathrm{V}=600\mu \mathrm{C}$