Question

A parallel plate capacitor of capacity $C_{0}$ is charged to a potential $V_{0}$. (i) The energy stored in the capacitor when the battery is disconnected and the plate separation is doubled is $E_{1}$. (ii) The energy stored in the capacitor when the charging battery is kept connected and the separation between the capacitor plates doubled is $E_{2}$. The ratio $E_{1} / E_{2}$ is________.

Answer 1

Initial energy: $E=\frac{1}{2} C_{0} V_{0}^{2}$
(i) $E_{1}=2 E$, because charge remains same and capacitance becomes half.
(ii) $E_{2}=E / 2$, because potential remains same and capacitance becomes half.
$\Rightarrow \frac{E_{1}}{E_{2}}=4$