Question 156 of 180

A parallel plate capacitor of capacitance $2F$ is charged to a potential $V$. The energy stored in the capacitor is ${E}_{1}$. The capacitor is now connected to another uncharged identical capacitor in parallel combination. The energy stored in the combination is ${E}_{2}$. The ratio $\dfrac{{E}_{2}}{{E}_{1}}$ is

A$2:1$B$2:3$C$1:2$D$1:4$

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Answer: (C) $1:2$

The charge on the plates of the first capacitor when connected against the potential difference $V$ is given by
$\begin{matrix}Q & = & CV \\ & = & 2V\end{matrix}$
When both the capacitors are connected, from the conservation of charge, it can be written that
$\begin{matrix}2V & = & 2{V}^{'}+2{V}^{'} \\ & \Rightarrow & {V}^{'}=\left(\dfrac{1}{2}V\right)\end{matrix}$
where, $V'$ is the new potential difference across each capacitor.
The formula to calculate the energy stored in the first capacitor is given by
$\begin{matrix}{E}_{1} & = & \dfrac{1}{2}\times C\times {V}^{2} \\ & = & \dfrac{1}{2}\times 2\times {V}^{2} \\ & = & {V}^{2}...\left(1\right)\end{matrix}$
For the second case, the energy stored in the combination of capacitor is given by
$\begin{matrix}{E}_{2} & = & \dfrac{1}{2}nCV{'}^{2} \\ & = & \dfrac{1}{2}\times 2\times \dfrac{{V}^{2}}{4}\times 2 \\ & = & \left(\dfrac{{V}^{2}}{2}\right)...\left(2\right)\end{matrix}$
Divide equation (2) by equation (1) to obtain the required ratio of the stored energy.
$\begin{matrix}\dfrac{{E}_{2}}{{E}_{1}} & = & \dfrac{\dfrac{{V}^{2}}{2}}{{V}^{2}} \\ & = & \left(\dfrac{1}{2}\right)\end{matrix}$