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

Vedclass · Accepted Answer

$20$

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(B) The energy stored in a capacitor is given by $U = \frac{Q^2}{2C}$,where $Q$ is the charge and $C$ is the capacitance.Let the initial charge be $Q$. The initial energy is $U_1 = \frac{Q^2}{2C}$.When the charge is increased by $2 \, C$,the new charge is $Q' = Q + 2$.The new energy is $U_2 = \frac{(Q+2)^2}{2C}$.Given that the energy increases by $21 \%$,the new energy is $U_2 = U_1 + 0.21 U_1 = 1.21 U_1$.Substituting the expressions for $U_1$ and $U_2$: $\frac{(Q+2)^2}{2C} = 1.21 \left( \frac{Q^2}{2C} \right)$.Canceling $\frac{1}{2C}$ from both sides,we get $(Q+2)^2 = 1.21 Q^2$.Taking the square root of both sides: $Q + 2 = \sqrt{1.21} Q$.$Q + 2 = 1.1 Q$.$0.1 Q = 2$.$Q = \frac{2}{0.1} = 20 \, C$.

If the charge on a capacitor is increased by $2 \, C$,the energy stored in it increases by $21 \%$. The original charge on the capacitor (in coulomb) is

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