If the charge on a capacitor is increased by $2$ coulomb, the energy stored in it increases by $21\%$. The original charge on the capacitor is....$C$
Medium
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Percentage increase in energy of capacitor $=\frac{\frac{q_{f}^{2}}{2 C}-\frac{q_{i}^{2}}{2 C}}{\frac{q_{i}^{2}}{2 C}} \times 100=21$ and $q_{f}-q_{i}=2$
Solving, we get $q_{i}=20 C$
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