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

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

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$20$

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(B) The energy stored in a charged body (like a capacitor) is given by $E = \frac{q^2}{2C}$,where $q$ is the charge and $C$ is the capacitance. Thus,$E \propto q^2$.Let the original charge be $q_1 = q$ and the original energy be $E_1 = E$.When the charge is increased by $2 \ C$,the new charge is $q_2 = q + 2$.The new energy $E_2$ increases by $21 \%$,so $E_2 = E + 0.21E = 1.21E$.Using the ratio: $\frac{E_2}{E_1} = \left(\frac{q_2}{q_1}\right)^2$.Substituting the values: $\frac{1.21E}{E} = \left(\frac{q+2}{q}\right)^2$.$1.21 = \left(\frac{q+2}{q}\right)^2$.Taking the square root on both sides: $1.1 = \frac{q+2}{q}$.$1.1q = q + 2$.$0.1q = 2$.$q = \frac{2}{0.1} = 20 \ C$.

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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