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A charge of $40\,\mu\text{C}$ is given to a capacitor of capacitance $10\,\mu\text{F}$. The stored energy in ergs is

1. $8\times 10^{-4}$
2. 800
3. 80
4. 8000

Correct answer: 800

Solution

Using $V=Q/C$, we get $V=40/10=4$ V. Then $U=\frac{1}{2}QV=\frac{1}{2}\times 40\times 10^{-6}\times 4=80\times 10^{-6}$ J, which is $800$ ergs. So the correct answer is 800.

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