Question

A parallel plate capacitor has a uniform electric field E in the space between the plates. If the distance between the plates is d and area of each plate is A, the eneigy stored in the capacitor is

• A. $ \frac{1}{2} \varepsilon_0 E^2 $
• B. $ \frac{E^2Ad}{\varepsilon_0} $
• C. $ \frac{1}{2} \varepsilon_0 E^2 Ad $
• D. $\varepsilon_0 EAd $

Answer 1

The Correct Option is C

To find the energy stored in a parallel plate capacitor, we need to derive the potential energy formula using the given parameters.

The basic formula for the energy stored in a capacitor is given by:

U = \frac{1}{2} C V^2

Here, C is the capacitance, and V is the potential difference between the plates of the capacitor.

The capacitance C for a parallel plate capacitor is given by:

C = \frac{\varepsilon_0 A}{d}

Where:

• \varepsilon_0 is the permittivity of free space.
• A is the area of each plate.
• d is the distance between the plates.

The electric field E between the plates is related to the potential difference V by:

E = \frac{V}{d} or V = E \cdot d

Substitute the expressions for C and V in the energy formula:

U = \frac{1}{2} \left(\frac{\varepsilon_0 A}{d}\right) \left(Ed\right)^2

Simplify the expression:

U = \frac{1}{2} \varepsilon_0 A d E^2

Thus, the energy stored in the capacitor is \frac{1}{2} \varepsilon_0 E^2 Ad.

Conclusion:

The correct answer is: \frac{1}{2} \varepsilon_0 E^2 Ad.