Question

The separation between the plates of a pre-charged parallel plate capacitor is increased. Which of the following quantities will change?

• A. Charge on the capacitor
• B. Potential difference across the capacitor
• C. Energy of the capacitor
• D. Energy density between the plates of the capacitor.

Answer 1

Answer: (B), (C)

The correct options are
B Potential difference across the capacitor
C Energy of the capacitor

Capacitance $C=\frac{{ϵ}_{0}A}{d}$
as $d$ increases, $C$ decreases.

Since the capacitor is pre-charged (charging battery removed) the charge remains the same.
$Q=CV$
$\Rightarrow CV=\text{constant}$
As $C$ decreases, $V$ should increase (option $\left(B\right)$ is correct)

Energy stored in the capacitor
$U=\frac{Q^2}{2C}$
as $C$ decreases, $U$ increases.(option $\left(C\right)$ is correct)

Energy density between the plates of the capacitor.
$E_d=\frac{U}{Ad}$
$E_d=\frac{Q^2}{2CAd}$
$E_d=\frac{Q^2}{2{ϵ}_0{A}^2}(\because C=\frac{{ϵ}_{0}A}{d})$
$E_d=\text{constant}$ (As $Q$ is constant)