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Q.
A capacitor of capacitance $ 6\,\mu F $ charged up to $100\, V$. The energy stored in the capacitor is:
BHUBHU 2003
Solution:
The total work done in charging the capacitor $(C)$ from the uncharged state to the final charge $q$ will be equal to energy stored in the capacitor, given by
$U=\frac{1}{2} \frac{q^{2}}{C}=\frac{1}{2} C V^{2}$
Where $q=C V$ and $V$ is potential difference.
Given, $C=6 \times 10^{-6}\, F, $
$V=100 $ volt
$\therefore U=\frac{1}{2} \times 6 \times 10^{-6} \times(100)^{2}$
$=3 \times 10^{-2} \,J$
$ U=0.03 \,J$
Note: This energy resides in the electric field created between the plates of the charged capacitor.