If the charge on the capacitor is increased by 3 C , the energy stored in it increases by 21% . The original charge on the capacitor is

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The initial energy stored in the capacitor is $U = \frac{Q ^{2}}{2 C}$ , where $Q$ is the original charge and $C$ is the capacitance.

When the charge is increased by $3 C$ , the new charge is $Q^{'} = Q + 3$ .

The new energy stored is $U^{'} = \frac{( Q + 3 ) ^{2}}{2 C}$ .

Given that the energy increases by $21%$ , $U^{'} = 1.21 U$ .

$\frac{( Q + 3 ) ^{2}}{2 C} = 1.21 \frac{Q ^{2}}{2 C}$

$(Q + 3)^{2} = 1.21 Q^{2}$

Taking the square root on both sides:

$Q + 3 = 1.1 Q$

$0.1 Q = 3$

$Q = 30 C$

Answer: 30 C

Physics - Capacitor Question with Solution | TestHub

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