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(B) The capacitance of a cylindrical capacitor is given by $C = \frac{2\pi \varepsilon_0 L}{\ln(b/a)}$.The energy stored in the capacitor is $U = \fra

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become double.

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(B) The capacitance of a cylindrical capacitor is given by $C = \frac{2\pi \varepsilon_0 L}{\ln(b/a)}$.The energy stored in the capacitor is $U = \frac{1}{2} \frac{Q^2}{C}$.Substituting the expression for $C$,we get $U = \frac{1}{2} \frac{Q^2 \ln(b/a)}{2\pi \varepsilon_0 L}$.Since $\ln(b/a)$,$\pi$,and $\varepsilon_0$ are constants,the energy $U$ is proportional to $\frac{Q^2}{L}$,i.e.,$U \propto \frac{Q^2}{L}$.Let the new charge be $Q' = 2Q$ and the new length be $L' = 2L$.The new energy $U'$ is proportional to $\frac{(Q')^2}{L'} = \frac{(2Q)^2}{2L} = \frac{4Q^2}{2L} = 2 \left( \frac{Q^2}{L} \right)$.Therefore,the energy becomes $2$ times the original energy.

$A$ cylindrical capacitor has charge $Q$ and length $L$. If both the length and the charge are doubled (keeping other parameters constant),the energy stored in the capacitor will:

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