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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 $q_i = q$ and the final charge be $q_f = q + 2$.The initial energy is $U_i = \frac{q^2}{2C}$ and the final energy is $U_f = \frac{(q+2)^2}{2C}$.Given that the energy increases by $21\%$,we have $U_f = U_i + 0.21 U_i = 1.21 U_i$.Substituting the expressions for energy: $\frac{(q+2)^2}{2C} = 1.21 	imes \frac{q^2}{2C}$.Canceling $\frac{1}{2C}$ from both sides: $(q+2)^2 = 1.21 q^2$.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,$q = \frac{2}{0.1} = 20 \ C$.

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 is....$C$

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