$=\frac{\text { Energy stored in capacitor }}{\text { Workdone by the battery }}=\frac{\frac{1}{2} C V^{2}}{C e^{2}}$
where $C=$ Capacitance of capacitor
$V=$ Potential difference
$e=\mathrm{emf}$ of battery
$=\frac{\frac{1}{2} C e^{2}}{C e^{2}}=\frac{1}{2} \quad(\because V=e)$





