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 :

10 C

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20 C

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30 C

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40 C

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Solution

The correct option is B $20 C$
According to question
Percentage increase in energy of capacitor= $\frac{\frac{q_{f}^{2}}{2 C} - \frac{q_{i}^{2}}{2 C}}{\frac{q_{i}^{2}}{2 C}} \times 100 = 21$ ... (i) and $q_{f} - q_{i} = 2$ ...(ii)

From (i) and (ii)

$4 (q_{i} + 1) = 0.21 (q_{i})^{2}$

On solving:

Solving, we get $q_{i} = 20 C$

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