Question

Prove that the electric potential energy per unit volume of a charged condenser is \(\tfrac{1}{2}\varepsilon_0 E^2\), where symbols have their usual meaning.

Hint:

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

Answer 1

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

Step 2: Take a parallel plate capacitor.
For a parallel plate capacitor of plate area \(A\) and plate separation \(d\) with vacuum (or air) between the plates,
\[ C = \frac{\varepsilon_0 A}{d} \]

Step 3: Relate potential difference to the field.
The field between the plates is uniform and equal to
\[ E = \frac{V}{d} \quad\Rightarrow\quad V = E\,d \]

Step 4: Substitute in the energy expression.
\[ U = \frac{1}{2}CV^2 = \frac{1}{2}\left(\frac{\varepsilon_0 A}{d}\right)(E\,d)^2 \]
\[ U = \frac{1}{2}\cdot\frac{\varepsilon_0 A}{d}\cdot E^2 d^2 = \frac{1}{2}\varepsilon_0 A d\,E^2 \]

Step 5: Divide by the volume.
The electric field exists only in the region between the plates, whose volume is
\[ \text{Volume} = A\,d \]
Hence the energy per unit volume (energy density) is
\[ u = \frac{U}{A d} = \frac{\tfrac{1}{2}\varepsilon_0 A d\,E^2}{A d} = \frac{1}{2}\varepsilon_0 E^2 \]
This proves the required result.

\[\boxed{u = \dfrac{1}{2}\varepsilon_0 E^2}\]