Question

A parallel plate capacitor, at a capacity $100\mu F$, is charged by a battery at $50V$. The battery remains connected and if the plates of the capacitor are separated so that the distance between them is reduced to half of the original distance, the additional energy given by the battery to the capacitor in $J$ is:

• A. $125\times {10}^{-3}$
• B. $12.5\times {10}^{-3}$
• C. $1.25\times {10}^{-3}$
• D. $0.125\times {10}^{-3}$

Answer 1

Answer: (A)

$125\times {10}^{-3}$

Initial energy:
$U_1=\frac{1}{2}C{V}^2$$=\frac{1}{2}\times 100\times {10}^{-6}\times (50{)}^2$$=12.5\times {10}^{-2}J$

When the battery is kept connected, voltage across the capacitor does not change.
We know: $C=\frac{{ϵ}_{0}A}{d}=100\mu F$
And when distance is halved:
$C^{\prime }=\frac{2{ϵ}_{0}A}{d}$$=200\mu F$

So, the final energy:

$U_2=\frac{1}{2}{C}^{\prime }{V}^2$$=\frac{1}{2}\times 200\times {10}^{-6}\times (50{)}^2$$=25\times {10}^{-2}J$

$\therefore$ additional energy given = $U_2-U_1$$=(25\times {10}^{-2}-12.5\times {10}^{-2})J$$=12.5\times {10}^{-2}J$