Question

A parallel plates capacitor is charged and the charging battery is then disconnected. If the plates of the capacitor are moved farther apart by means of insulating handles, then

• A. the charge on the capacitor increases
• B. the voltage across the plates increases
• C. the capacitance increases
• D. the electrostatic energy stored in the capacitor increases.

Answer 1

Answer: (B), (D)

the electrostatic energy stored in the capacitor increases.

Charge on plate is $q$
Capacitance of parallel plate capacitor is
$C=\frac{{ϵ}_{0}A}{d}$
we know that
$q=CV$

$\Rightarrow V=\frac{q}{C}$
Potential energy stored in capacitor is $U=\frac{q^2}{2C}$

Given that distance between capacitors is increased to $d^{\prime }$
Charge on plate remains $q$ as charging battery is removed. The charge on the capacitor remains constant.

New capacitance
$C^{\prime }=\frac{{ϵ}_{0}A}{d^{\prime }}<C$

Capacitance decreases as the plates of capacitor are moved farther apart.
The options (a) and (c) are incorrect.

Potential across the plates $V=\frac{q}{C^{\prime }}>\frac{q}{C}$
Hence $V$ increases.
Option (b) is correct.

Energy stored in the capacitor $U=\frac{1}{2}\frac{q^2}{C^{\prime }}>\frac{1}{2}\frac{q^2}{C}$
$\therefore U$ increases.
Option (d) is correct.