Question

A capacitor of capacitance $6\mu F$ charged up to $100V$. The energy stored in the capacitor is:

• A. 0.6 J
• B. 0.06 J
• C. 0.03 J
• D. 0.3 J

Answer 1

Answer: (C)

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 $q=CV$ and $V$ is potential difference.
Given, $C=6\times 10^{-6}F,$
$V=100$ volt
$\therefore U=\frac{1}{2}\times 6\times 10^{-6}\times (100{)}^2$
$=3\times 10^{-2}J$
$U=0.03J$
Note: This energy resides in the electric field created between the plates of the charged capacitor.