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 \( \epsilon_0 \) is the permittivity of free space between the plates, then the potential energy stored in the capacitor is:

• A. \( \frac{1}{2} \epsilon_0 E^2 A d \)
• B. \( \frac{3}{4} \epsilon_0 E^2 A d \)
• C. \( \frac{1}{4} \epsilon_0 E^2 A d \)
• D. \( \epsilon_0 E^2 A d \)

Hint:

For parallel plate capacitors, remember the formula for the capacitance \( C = \frac{\epsilon_0 A}{d} \) and the relationship between the electric field and potential difference \( V = E d \). These are essential in calculating the potential energy stored in the capacitor.

Answer 1

The Correct Option is A

The potential energy stored in a parallel plate capacitor is determined by starting with the fundamental formula for capacitor energy. The energy \(U\) stored is expressed as:

\(U = \frac{1}{2}CV^2\)

Here, \(C\) represents the capacitance of the capacitor, and \(V\) is the potential difference across its plates.

The capacitance \(C\) for a parallel plate capacitor is defined by:

\(C = \frac{\epsilon_0 A}{d}\)

In this equation, \(\epsilon_0\) is the permittivity of free space, \(A\) is the area of the plates, and \(d\) is the distance between the plates.

The electric field \(E\) established between the capacitor plates is given by:

\(E = \frac{V}{d}\)

By substituting \(V = Ed\) into the energy formula, we obtain:

\(U = \frac{1}{2}C(Ed)^2\)

Further substitution of the expression for \(C\) yields:

\(U = \frac{1}{2}\left(\frac{\epsilon_0 A}{d}\right)(Ed)^2\)

Simplification of this expression results in:

\(U = \frac{1}{2} \epsilon_0 E^2 A d\)

This result is consistent with the provided option \(\frac{1}{2} \epsilon_0 E^2 A d\), validating it as the correct answer.