Question

A parallel plate air capacitor having area of each plate '\(A\)' and the distance between the plates '\(d\)' has uniform electric field E in the space between the plates. The energy stored in the capacitor is (\(ε_0\) = permittivity of free space.)

• A. \(\frac{1}{2}ε_0E^2\)
• B. \(\frac{1}{2}Aε_0E\)
• C. \(\frac{1}{2}Aε_0E^2d\)
• D. \(\frac{1}{2}Aε_0^2E\cdot d\)

Hint:

Energy density is (1/2) e0 E squared; multiply by volume A d.

Answer 1

The Correct Option is C

Step 1: Use charge form:
$U = \frac{Q^2}{2C}$. With $Q = \sigma A = \varepsilon_0EA$ and $C = \frac{\varepsilon_0A}{d}$:

Step 2: Compute:
$U = \frac{(\varepsilon_0EA)^2}{2\varepsilon_0A/d} = \frac{\varepsilon_0^2E^2A^2d}{2\varepsilon_0A} = \frac12\varepsilon_0E^2Ad$.

Step 3: Check units:
$\varepsilon_0E^2$ is energy per volume, times $Ad$ (volume) gives energy.

Final Answer:
The energy is (1/2) A e0 E squared d, option (C). \[ \boxed{\frac{1}{2}A\varepsilon_0E^2d} \]