If the charge on the capacitor is increased b | vedclass

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(C) The energy stored in a capacitor is given by $U = \frac{q^2}{2C}$.Let the original charge be $q$. The original energy is $U = \frac{q^2}{2C}$.When

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$20$

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(C) The energy stored in a capacitor is given by $U = \frac{q^2}{2C}$.Let the original charge be $q$. The original energy is $U = \frac{q^2}{2C}$.When the charge is increased by $2 \ C$,the new charge becomes $(q + 2) \ C$.The new energy $U'$ is $21 \%$ more than the original energy,so $U' = U + 0.21U = 1.21U$.Substituting the expressions for energy: $\frac{(q + 2)^2}{2C} = 1.21 	imes \frac{q^2}{2C}$.Canceling $2C$ 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: $1.1q - q = 2$,which gives $0.1q = 2$.Therefore,$q = \frac{2}{0.1} = 20 \ C$.

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

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