Rankers Physics

Capacitors: Practice Problem & Solution

Energy per unit volume for a capacitor having area $A$ and separation $d$ kept at potential difference $V$ is given by: (2001)
$\frac{1}{2} \varepsilon_{0} \frac{V^{2}}{d^{2}}$
$\frac{1}{2 \varepsilon_{0}} \frac{V^{2}}{d^{2}}$
$\frac{1}{2} C V^{2}$
$\frac{Q^{2}}{2 C}$

Short Solution (Best for NEET):

To solve this problem, we apply the core principles of Capacitors. Understanding the underlying formula is key to arriving at the correct answer below:

Energy density $u = \frac{1}{2} \varepsilon_{0} E^{2}$. For a parallel plate capacitor, electric field $E = \frac{V}{d}$. Therefore, $u = \frac{1}{2} \varepsilon_{0} \frac{V^{2}}{d^{2}}$.

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