2001 AIPMT Electrostatic Potential and CapacitanceEnergy Stored in a Capacitor Easy Question 6 of 6 in Energy Stored in a Capacitor

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

A $\dfrac{1}{2}\varepsilon_0\dfrac{V^2}{d^2}$
B $\dfrac{1}{2\varepsilon_0}\dfrac{V^2}{d^2}$
C $\dfrac{1}{2}CV^2$
D $\dfrac{Q^2}{2C}$

The answer is A.

Solution

Energy stored: $U = \dfrac{1}{2}CV^2$ with $C = \dfrac{\varepsilon_0A}{d}$
Volume between the plates $= Ad$
Energy density $u = \dfrac{U}{Ad} = \dfrac{\frac{1}{2}\frac{\varepsilon_0A}{d}V^2}{Ad} = \dfrac{1}{2}\varepsilon_0\dfrac{V^2}{d^2}$
(Since $E = \dfrac{V}{d}$, this is $\dfrac{1}{2}\varepsilon_0E^2$.)
$\dfrac{1}{2}CV^2$ and $\dfrac{Q^2}{2C}$ are the total energy, not the energy per unit volume.

Concept Energy density of electric field

Energy Stored in a Capacitor in past papers

6 questions from this chapter have appeared across 6 exam years.