Question

A parallel plate capacitor is charged with a battery of emf $V$ volts. After charging, the battery is disconnected and the distance between the plates is increased to $1.5$ times the initial value. The new potential across the plates of the capacitor will become

$[$0.77 Mark$]$

• A. $V$
• B. $2V$
• C. $1.5V$
• D. $2.8V$

Answer 1

The correct option is C $1.5V$

On disconnecting the battery, charge on the capacitor would remain constant.
When the distance between the plates of a parallel plate capacitor is increased by $1.5$ times, then the capacitance of the parallel plate capacitor decreases by $1.5$ times.
$C=\frac{{ϵ}_{o}A}{d}$
$C^{{}^{\prime }}=\frac{{ϵ}_{o}A}{1.5d}=\frac{C}{1.5}$

As,
$V^{\prime }=\frac{Q}{C^{{}^{\prime }}}=\frac{Q}{\left(\frac{C}{1.5}\right)}=1.5\frac{Q}{C}=1.5V$

Hence, option $\left(\text{c}\right)$ is the right answer.