When an additional charge of 2 text{ C} is gi | vedclass

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(D) The energy stored in a capacitor is given by $U = \frac{q^2}{2C}$.Let the initial charge be $q_1 = q$ and the final charge be $q_2 = q + 2 	ext{

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$20$

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(D) The energy stored in a capacitor is given by $U = \frac{q^2}{2C}$.Let the initial charge be $q_1 = q$ and the final charge be $q_2 = q + 2 	ext{ C}$.The initial energy is $U_1 = \frac{q^2}{2C}$ and the final energy is $U_2 = \frac{(q+2)^2}{2C}$.Given that the energy increases by $21 \%$,we have $U_2 = U_1 + 0.21 U_1 = 1.21 U_1$.Substituting the expressions for $U_1$ and $U_2$:$\frac{(q+2)^2}{2C} = 1.21 	imes \frac{q^2}{2C}$.Canceling $\frac{1}{2C}$ from both sides,we get $(q+2)^2 = 1.21 q^2$.Taking the square root of both sides: $q + 2 = 1.1 q$.Rearranging the terms: $1.1 q - q = 2$,which gives $0.1 q = 2$.Therefore,$q = \frac{2}{0.1} = 20 	ext{ C}$.

When an additional charge of $2 \text{ C}$ is given to a capacitor,the energy stored in it increases by $21 \%$. The original charge of the capacitor is: (in $\text{ C}$)

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