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CGP EDU Academic Team
Published on: September 11, 2026
A charge of $40 \mu C$ is given to a capacitor having capacitance $C = 10 \mu F$ . The stored energy in ergs is
Text Solution
Verified by ExpertsThe correct answer is:
B
$U = \frac{Q^2}{2C} = \frac{\left(40 \times 10^{-6}\right)^2}{2 \times 10^{-6} \times 10} = \frac{16 \times 10^{-10}}{2 \times 10^{-5}} = 8 \times 10^{-5} J$
$= 8 \times 10^{-5} \times 10^{7} = 800 \mathrm{erg}$
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