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Q.
When an additional charge of $2C$ is given to a capacitor, energy stored in it is increased by $21\%$. The original charge of the capacitor is
KCETKCET 2013Electrostatic Potential and Capacitance
Solution:
Energy stored by a capacitor, $U=\frac{1}{2} \frac{q^{2}}{C}$
So, here $U_{1}=\frac{1}{2} \frac{q_{1}^{2}}{C} ;\,\,\, U_{2}=\frac{1}{2} \frac{q_{2}^{2}}{C}$
$\therefore \frac{U_{1}}{U_{2}}=\left(\frac{q_{1}}{q_{2}}\right)^{2}=\left(\frac{q}{q+2}\right)^{2} $
Given, $\frac{U_{2}-U_{1}}{U_{1}}=0.21$
or $\frac{U_{2}}{U_{1}}=1.21 \Rightarrow \left(\frac{q}{q+2}\right)^{2}=1.21$
On solving we, get
$q=20 \,C$