Q132 / 150 · MHT CET 2011 · Physics
Energy per unit volume for a capacitor having area $A$ and separation $d$ kept at potential difference $V$ is given by
- Correct answerA$\frac{1}{2} \varepsilon_{0} \frac{V^{2}}{d^{2}}$
- B$\frac{1}{2 \varepsilon_{0}} \frac{V^{2}}{d^{2}}$
- C$\frac{1}{2} C V^{2}$
- D$\frac{Q^{2}}{2 C}$
Solution
Energy density $=\frac{1}{2} \varepsilon_{0} E^{2}$
Since
$$
E=\frac{V}{d}
$$
Therefore energy $=\frac{1}{2} \varepsilon_{0}\left(\frac{V^{2}}{d^{2}}\right)$
Hence, the correct answer is option: 'A'.