Question

A parallel plate capacitor is charged by a battery to a potential difference $V$. It is disconnected from battery and then connected to another uncharged capacitor of the same capacitance. Calculate the ratio of the energy stored in the combination to the initial energy on the single capacitor.

Answer 1

Let charge $q$ be transferred from one capacitor to the other. Then equating potential difference across the two capacitors,

$\frac{CV-q}{C}=\frac{q}{C}$


$\Rightarrow q=\frac{CV}{2}$

Now, Energy stored in the combination,

$U_f=2\frac{1}{2}\frac{q^2}{C}=\frac{C{V}^2}{4}$

Initial energy, $U_i=\frac{1}{2}C{V}^2$


$\Rightarrow \frac{U_f}{U_i}=\frac{C{V}^2}{4}\frac{2}{C{V}^2}=\frac{1}{2}$