If the charge on the capacitor is increased b | vedclass

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Question

(A) Let the original charge on the capacitor be $Q$ and the capacitance be $C$. The energy stored in the capacitor is given by $U = \frac{Q^2}{2C}$.Wh

Vedclass · Accepted Answer

$20$

Vedclass · Answer

(A) Let the original charge on the capacitor be $Q$ and the capacitance be $C$. The energy stored in the capacitor is given by $U = \frac{Q^2}{2C}$.When the charge is increased by $2 	ext{ C}$, the new charge becomes $Q' = Q + 2$.The new energy stored is $U' = \frac{(Q + 2)^2}{2C}$.Given that the energy increases by $21 \%$, we have $U' = U + 0.21U = 1.21U$.Substituting the expressions for $U$ and $U'$, we get $\frac{(Q + 2)^2}{2C} = 1.21 	imes \frac{Q^2}{2C}$.Canceling $\frac{1}{2C}$ from both sides, we get $(Q + 2)^2 = 1.21Q^2$.Taking the square root of both sides, $Q + 2 = 1.1Q$.Rearranging the terms, $0.1Q = 2$, which gives $Q = \frac{2}{0.1} = 20 	ext{ C}$.

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

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