A parallel plate capacitor having plate area $A$ and separation $d$ is charged to a potential difference $V$…
A parallel plate capacitor having plate area $A$ and separation $d$ is charged to a potential difference $V$. The charging battery is disconnected and the plates are pulled apart to four times the initial separation. The work required to increase the distance between plates is:
$\frac{\varepsilon_0 A V^2}{4 d}$
$\frac{2 \varepsilon_0 A V^2}{4 d}$
$\frac{\varepsilon_0 A V^2}{3 d}$
$\frac{3 \varepsilon_0 A V^2}{2 d}$
Solution
Initial energy: $\frac{q^2}{2 C}$
The new capacitance is: $C_n=\frac{C}{4}$
Final energy after the change in capacitance is: $\frac{4 q^2}{2 C}$
Therefore, magnitude of work done is equal to the change in potential energy,
$W=\frac{2 q^2}{C}-\frac{q^2}{2 C}=\frac{3 C V^2}{2}=\frac{3 \varepsilon_0 A V^2}{2 d}$