Question

The capacitance of a capacitor is $10F$. The potential difference on it is $50V$. If the distance between its plate is halved, What will be the potential difference now?

• A. $100V$
• B. $50V$
• C. $25V$
• D. $75V$

Answer 1

Answer: (C)

Given, $C=10F$
$V=50V$
If the distance between the plate is halved.
$C=\frac{A{ϵ}_0}{d}$
$C^{\prime }=\frac{2A{ϵ}_0}{d}=2C$
Since energy is not dissipated, it remain constant.
The surface charge density $\left(\sigma \right)$ will remain constant.

$E=\frac{\sigma }{{ϵ}_0}$[$\therefore E$ is also constant]

$V=E\times d$
$\Rightarrow V^{\prime }=E{d}^{\prime }$
$\Rightarrow V^{\prime }=\frac{V}{2}=\frac{50}{2}=25$
This shows that V will also get halved.