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 given by the formula: \[ U = \frac{1}{2} C V^2, \] where \( C \) is the capacitance of the capacitor and \( V \) is the potential difference across the plates. 
The capacitance \( C \) of a parallel plate capacitor is given by: \[ C = \frac{\epsilon_0 A}{d}, \] where \( A \) is the area of each plate, \( \epsilon_0 \) is the permittivity of free space, and \( d \) is the separation between the plates. 
The potential difference \( V \) across the plates is related to the electric field \( E \) by: \[ V = E d, \] where \( E \) is the electric field. Now substitute these into the formula for potential energy: \[ U = \frac{1}{2} \left( \frac{\epsilon_0 A}{d} \right) (E d)^2. \] Simplifying: \[ U = \frac{1}{2} \epsilon_0 E^2 A d. \] Thus, the potential energy stored in the capacitor is: \[ \boxed{\frac{1}{2} \epsilon_0 E^2 A d}. \]