Q109 / 125 · BITSAT 2011 · Physics
If the potential of a capacitor having capacity $6 \mu \mathrm{F}$ is increased from $10 \mathrm{~V}$ to $20 \mathrm{~V}$, then increase in its energywill be
- A$4 \times 10^{-4} \mathrm{~J}$
- B$4 \times 10^{-6} \mathrm{~J}$
- Correct answerC$9 \times 10^{-4} \mathrm{~J}$
- D$12 \times 10^{-6} \mathrm{~J}$
Solution
Capacitance of capacitor (C) $=6 \mu \mathrm{F}$ $=6 \times 10^{-6} \mathrm{~F}$; Initial potential $\left(\mathrm{V}_{1}\right)=10 \mathrm{~V}$ and
final potential $\left(\mathrm{V}_{2}\right)=20 \mathrm{~V}$.
The increase in energy $(\Delta \mathrm{U})$
$$
\begin{array}{l}
=\frac{1}{2} \mathrm{C}\left(\mathrm{V}_{2}^{2}-\mathrm{V}_{1}^{2}\right) \\
=\frac{1}{2} \times\left(6 \times 10^{-6}\right) \times\left[(20)^{2}-(10)^{2}\right] \\
=\left(3 \times 10^{-6}\right) \times 300=9 \times 10^{-4} \mathrm{~J} .
\end{array}
$$
Hence, the correct answer is option: 'C'.