Potential, $E^{\prime}=E\left(1-e^{-t / R C}\right)$

$\therefore$ Charge, $Q=C E\left(1-e^{\frac{-t}{R C}}\right) \quad\left(\because Q=C E^{\prime}\right)$

As capacitor is charged to $\frac{E}{2}$.

$\Rightarrow Q =\frac{C E}{2}$

$\therefore \frac{C E}{2}=C E\left(1-e^{\frac{-t}{R C}}\right)$

$\Rightarrow \frac{1}{2}=e^{\frac{-t}{R C}} \text { or } t=\frac{R C}{\ln 2}$

Work done by battery,

Work done by battery,

$W=Q \times \Delta V=\frac{C E}{2} \times E=\frac{C E^{2}}{2}$

Heat dissipated $=\int \limits_{0}^{R C / ln 2} i R d t$

$=\frac{E^{2}}{R} \int \limits_{0}^{R C / \ln 2} e^{\frac{-2 t}{R C}} d t=\frac{3}{4}\left(\frac{C E^{2}}{2}\right)$

$\therefore$ Work done $\frac{\left(\frac{C E^{2}}{2}\right)}{\frac{3}{8} C E^{2}}$

$=\frac{4}{3}$ or $4: 3$