A parallel plate capacitor of capacitance 2   F is charged to a potential V . The energy stored in the…

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

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

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

$\begin{array}{rcl}2V& =& 2V^{\text{'}}+2V^{\text{'}}\\ & \Rightarrow & V^{\text{'}}=\left(\frac{1}{2}V\right)\end{array}$

where, $V\text{'}$ is the new potential difference across each capacitor.

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

$\begin{array}{rcl}{E}_1& =& \frac{1}{2}\times C\times V^2\\ & =& \frac{1}{2}\times 2\times V^2\\ & =& V^2 ...\left(1\right)\end{array}$

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

$\begin{array}{rcl}{E}_2& =& \frac{1}{2}nCV{\text{'}}^2\\ & =& \frac{1}{2}\times 2\times \frac{V^2}{4}\times 2\\ & =& \left(\frac{V^2}{2}\right) ...\left(2\right)\end{array}$

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

$\begin{array}{rcl}\frac{E_2}{E_1}& =& \frac{{\displaystyle \frac{V^2}{2}}}{V^2}\\ & =& \left(\frac{1}{2}\right)\end{array}$