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(A) The energy stored in a capacitor is given by $U = \frac{q^2}{2C}$,where $q$ is the charge and $C$ is the capacitance.Since $C$ is constant,$U \pro

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$10$

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(A) The energy stored in a capacitor is given by $U = \frac{q^2}{2C}$,where $q$ is the charge and $C$ is the capacitance.Since $C$ is constant,$U \propto q^2$.Let the initial charge be $q$ and the initial energy be $U_i = \frac{q^2}{2C}$.When the charge is increased by $2 \ C$,the new charge is $q' = q + 2$.The new energy is $U_f = \frac{(q+2)^2}{2C}$.Given that the energy increases by $44 \%$,the new energy is $U_f = U_i + 0.44 \, U_i = 1.44 \, U_i$.Substituting the expressions for $U_f$ and $U_i$:$\frac{(q+2)^2}{2C} = 1.44 	imes \frac{q^2}{2C}$.$(q+2)^2 = 1.44 \, q^2$.Taking the square root on both sides:$q + 2 = \sqrt{1.44} \, q$.$q + 2 = 1.2 \, q$.$0.2 \, q = 2$.$q = \frac{2}{0.2} = 10 \ C$.

If the charge on a capacitor is increased by $2 \ C$,the energy stored in it increases by $44 \%$. The original charge on the capacitor is (in $C$):

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