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(B) The energy stored in a capacitor is given by $U = \frac{Q^2}{2C}$.Since $C$ is constant,$U \propto Q^2$.Let the initial charge be $Q_1$ and the in

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$15$

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(B) The energy stored in a capacitor is given by $U = \frac{Q^2}{2C}$.Since $C$ is constant,$U \propto Q^2$.Let the initial charge be $Q_1$ and the initial energy be $U_1$.When the charge is increased by $3 \ C$,the new charge is $Q_2 = Q_1 + 3$.The new energy $U_2$ is $U_1 + 44\% 	ext{ of } U_1 = 1.44 \ U_1$.Using the proportionality $U \propto Q^2$,we have:$\frac{U_2}{U_1} = \left( \frac{Q_2}{Q_1} ight)^2$$1.44 = \left( \frac{Q_1 + 3}{Q_1} ight)^2$Taking the square root on both sides:$1.2 = \frac{Q_1 + 3}{Q_1}$$1.2 \ Q_1 = Q_1 + 3$$0.2 \ Q_1 = 3$$Q_1 = \frac{3}{0.2} = 15 \ C$.

If the charge on the capacitor is increased by $3 \ C$,the energy stored in it increases by $44 \%$. The original charge on the capacitor is (in $C$)

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