Question

A parallel plate capacitor of capacity $C_{\circ }$ is charged to a potential $V_{\circ }$ .The energy stored in the capacitor when the battery is disconnected and the plate separation is doubled is $E_1$ and the energy stored in the capacitor when the charging battery is kept connected and the separation between the capacitor plates is doubled is $E_2$. Then value of $\frac{E_1}{E_2}$ is:

• A. $4$
• B. $\frac{3}{2}$
• C. $2$
• D. $\frac{1}{2}$

Answer 1

The correct option is A $4$

I) When battery is disconnected charge on capacitor is constant
We know $C=\frac{{\epsilon }_{0}A}{d}$
when d is doubled $C_1=\frac{{\epsilon }_{0}A}{2d}$
$\therefore Now{C}_1=\frac{C}{2}$
$Q=C_1{V}_1$
$\Rightarrow V_1=2\frac{Q}{C}=2V$
$\therefore E_1=\frac{1}{2}{C}_1^0{V}_1^2=\frac{1}{2}\cdot \frac{C}{2}\cdot 4V^2=C{V}^2$
II) When battery is connected V does not change
Now $C_2=\frac{{\epsilon }_{0}A}{2d}$
$\therefore E_2=\frac{1}{2}{C}_2{V}_2^2=\frac{1}{2}\cdot \left(\frac{C}{2}\right)V^2=\frac{C{V}^2}{4.}$
$\therefore \frac{E_1}{E_2}=\frac{C{V}^2}{\frac{C{V}^2}{4}}=4$