A capacitor of capacitance $C$ is charged to a potential $V$ and then connected to an uncharged capacitor of capacitance $2C$. The loss of energy in the process is

Solution

1. The initial energy stored in the capacitor is $U_i=\frac{1}{2}C{V}^2$.
2. When connected to an uncharged capacitor of $2C$, the total charge $Q=CV$ is conserved, resulting in a common potential $$\begin{array}{rl}{V}^{\prime }& =\frac{CV}{C+2C}\\ & =\frac{V}{3}\end{array}$$.
3. The final energy stored in the combined system is $$\begin{array}{rl}{U}_f& =\frac{1}{2}(C+2C)(V^{\prime }{)}^2\\ & =\frac{1}{2}(3C)(\frac{V}{3}{)}^2\\ & =\frac{1}{6}C{V}^2\end{array}$$.
4. The energy loss is $$\begin{array}{rl}\Delta U& =U_i-U_f\\ & =\frac{1}{2}C{V}^2-\frac{1}{6}C{V}^2\\ & =\frac{1}{3}C{V}^2\end{array}$$.
5. Hence the answer is (A).