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)
Solution
<p>Let initially the charge is q so</p>
<p>${1 \over 2}{{{q^2}} \over C} = {U_i}$</p>
<p>And ${1 \over 2}{{{{(q + 2)}^2}} \over C} = {U_f}$</p>
<p>Given ${{{U_f} - {U_i}} \over {{U_i}}} \times 100 = 44$</p>
<p>${{{{(q + 2)}^2} - {q^2}} \over q} = .44$</p>
<p>$\Rightarrow q = 10C$</p>
About this question
Subject: Physics · Chapter: Electrostatics · Topic: Electric Potential and Capacitance
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