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: Understanding the Concept:
Energy stored in a capacitor is \(U = \frac12CV^2\). For a parallel plate capacitor \(C = \frac{\varepsilon_0A}{d}\) and \(V = Ed\).

Step 2: Substitute:
\[ U = \frac12\cdot\frac{\varepsilon_0A}{d}\cdot(Ed)^2 = \frac12\varepsilon_0AE^2d \]

Step 3: Interpretation:
This equals the energy density \(\frac12\varepsilon_0E^2\) multiplied by the volume \(Ad\) between the plates.

Step 4: Why the other options are wrong.
\(\frac12\varepsilon_0E^2\) is only the energy per unit volume. \(\frac12A\varepsilon_0E\) and \(\frac12A\varepsilon_0^2E d\) have wrong powers of \(E\) and \(\varepsilon_0\).

Final Answer:
The stored energy is \(\frac12A\varepsilon_0E^2d\), option (C). \[ \boxed{\frac{1}{2}A\varepsilon_0E^2d} \]