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A battery is used to charge a parallel plate capacitor till the potential difference between the plates becomes equal to the e.m.f of the battery. The ratio of the energy stored in the capacitor to the work done by the battery will be
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The correct answer is:
$\frac{1}{2}$
Energy stored in the capacitor, $\mathrm{U}=\frac{1}{2} \mathrm{qV}$ Work done by the battery, $\mathrm{W}=\mathrm{qV}$
$\therefore \frac{\mathrm{U}}{\mathrm{W}}=\frac{1}{2}$
$\therefore \frac{\mathrm{U}}{\mathrm{W}}=\frac{1}{2}$
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