Question

Prove that the energy density inside a charged capacitor is \( \tfrac{1}{2}\varepsilon_0 E^2 \).

Hint:

Use \( U=\tfrac12 CV^2 \) with \( C=\varepsilon_0 A/d \) and \( V=Ed \), then divide the stored energy by the volume \( A\,d \) between the plates.

Answer 1

Step 1: Energy stored in a capacitor.
The energy stored in a charged parallel plate capacitor of capacitance \( C \) charged to a potential difference \( V \) is
\[ U = \frac{1}{2}CV^2 \]

Step 2: Substitute the capacitor parameters.
For a parallel plate capacitor with plate area \( A \) and plate separation \( d \), the capacitance is
\[ C = \frac{\varepsilon_0 A}{d} \]
The potential difference is related to the uniform field \( E \) between the plates by
\[ V = E\,d \]

Step 3: Put these into the energy expression.
\[ U = \frac{1}{2}\left(\frac{\varepsilon_0 A}{d}\right)(E\,d)^2 = \frac{1}{2}\,\varepsilon_0\,\frac{A}{d}\,E^2 d^2 = \frac{1}{2}\varepsilon_0 E^2 (A\,d) \]

Step 4: Divide by the volume to get energy density.
The electric field exists in the space between the plates, whose volume is \( \text{Volume} = A\,d \). Energy density is energy per unit volume:
\[ u = \frac{U}{A\,d} = \frac{\tfrac{1}{2}\varepsilon_0 E^2 (A\,d)}{A\,d} \]

Step 5: Result.
The area and separation cancel, giving
\[\boxed{u = \frac{1}{2}\varepsilon_0 E^2}\]
which is independent of the plate dimensions and depends only on the field, proving the result.