Question

Consider a parallel plate capacitor of area \(A\) of each plate and separation \(d\) between the plates. If \(E\) is the electric field and \(\varepsilon_0\) is the permittivity of free space between the plates, then potential energy stored in the capacitor is:

• A. \(\displaystyle \frac{1}{2}\varepsilon_0 E^2Ad\)
• B. \(\displaystyle \frac{3}{4}\varepsilon_0 E^2Ad\)
• C. \(\displaystyle \frac{1}{4}\varepsilon_0 E^2Ad\)
• D. \(\displaystyle \varepsilon_0 E^2Ad\)

Hint:

Energy density in an electric field is \(\frac{1}{2}\varepsilon_0E^2\). Multiply it by volume \(Ad\) to get total energy stored.

Answer 1

The Correct Option is A

Concept:
The energy density of an electric field is given by: \[ u=\frac{1}{2}\varepsilon_0 E^2 \] where \(u\) is energy per unit volume.

Step 1: Find the volume between the capacitor plates.
For a parallel plate capacitor, \[ \text{Volume}=\text{Area}\times \text{separation} \] \[ V_{\text{volume}}=A \times d \] \[ V_{\text{volume}}=Ad \]

Step 2: Use energy density formula.
Energy stored is: \[ U=\text{Energy density}\times \text{Volume} \] \[ U=\left(\frac{1}{2}\varepsilon_0 E^2\right)(Ad) \] \[ U=\frac{1}{2}\varepsilon_0 E^2Ad \]

Step 3: Final conclusion.
Hence, the potential energy stored in the capacitor is: \[ \boxed{\frac{1}{2}\varepsilon_0 E^2Ad} \]