A parallel plate capacitor of capacitance 2F is charged to a...

The charge on the plates of the first capacitor when connected against the potential difference $V$ is given by

$\begin{cases}Q & = & CV \\ = & 2V\end{cases}$

When both the capacitors are connected, from the conservation of charge, it can be written that

$\begin{cases}2V & = & 2{V}^{'} + 2{V}^{'} \\ ⇒ & {V}^{'} = \left(\dfrac{1}{2}V\right)\end{cases}$

where, $V '$ is the new potential difference across each capacitor.

The formula to calculate the energy stored in the first capacitor is given by

$\begin{cases}{E}_{1} & = & \dfrac{1}{2} \times C \times {V}^{2} \\ = & \dfrac{1}{2} \times 2 \times {V}^{2} \\ = & {V}^{2} . . . \left(1\right)\end{cases}$

For the second case, the energy stored in the combination of capacitor is given by

$\begin{cases}{E}_{2} & = & \dfrac{1}{2}nCV{'}^{2} \\ = & \dfrac{1}{2} \times 2 \times \dfrac{{V}^{2}}{4} \times 2 \\ = & \left(\dfrac{{V}^{2}}{2}\right) . . . \left(2\right)\end{cases}$

Divide equation (2) by equation (1) to obtain the required ratio of the stored energy.

$\begin{cases}\dfrac{{E}_{2}}{{E}_{1}} & = & \dfrac{\dfrac{{V}^{2}}{2}}{{V}^{2}} \\ = & \left(\dfrac{1}{2}\right)\end{cases}$