JEE MAIN - Physics (2022 - 24th June Evening Shift - No. 14)
If the charge on a capacitor is increased by 2 C, the energy stored in it increases by 44%. The original charge on the capacitor is (in C)
10
20
30
40
Explanation
Let initially the charge is q so
$${1 \over 2}{{{q^2}} \over C} = {U_i}$$
And $${1 \over 2}{{{{(q + 2)}^2}} \over C} = {U_f}$$
Given $${{{U_f} - {U_i}} \over {{U_i}}} \times 100 = 44$$
$${{{{(q + 2)}^2} - {q^2}} \over q} = .44$$
$$ \Rightarrow q = 10C$$
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