Question

Energy per unit volume for a capacitor having area $ A $ and separation d kept at potential difference $ V $ is given by

• A. $ \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} CV^2 $
• D. $ \frac{Q^2}{2C} $

Answer 1

Answer: (A)

Energy density $=\frac{1}{2} \varepsilon_{0} E^{2}$
Since $\,\,\, E=\frac{V}{d}$
Therefore energy density $=\frac{1}{2} \varepsilon_{0}\left(\frac{V^{2}}{d^{2}}\right)$