If the charge on a body is increased by 2 C, | vedclass

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(B) The energy stored in a capacitor or a charged body is given by $E = \frac{q^2}{2C}$,where $q$ is the charge and $C$ is the capacitance.Since $E \p

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$20$

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(B) The energy stored in a capacitor or a charged body is given by $E = \frac{q^2}{2C}$,where $q$ is the charge and $C$ is the capacitance.Since $E \propto q^2$,we can write $\frac{E_2}{E_1} = \left(\frac{q_2}{q_1}\right)^2$.Given that the energy increases by $21 \%$,the new energy $E_2 = E_1 + 0.21 E_1 = 1.21 E_1$.Thus,$\frac{E_2}{E_1} = 1.21 = \left(\frac{11}{10}\right)^2$.Let the original charge be $q$. The new charge is $q_2 = q + 2$.Substituting these into the ratio equation: $\left(\frac{q+2}{q}\right)^2 = \left(\frac{11}{10}\right)^2$.Taking the square root on both sides: $\frac{q+2}{q} = \frac{11}{10}$.Cross-multiplying gives: $10(q + 2) = 11q$.$10q + 20 = 11q$.$q = 20 \ C$.

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

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