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:
Show Hint
- \( \frac{1}{2} \epsilon_0 E^2 A d \)
- \( \frac{3}{4} \epsilon_0 E^2 A d \)
- \( \frac{1}{4} \epsilon_0 E^2 A d \)
- \( \epsilon_0 E^2 A d \)
The Correct Option is A
Approach Solution - 1
To find the potential energy stored in a parallel plate capacitor, we start by considering the formula for the energy stored in a capacitor. The energy \(U\) stored in a capacitor is given by:
\(U = \frac{1}{2}CV^2\)
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 \(\epsilon_0\) is the permittivity of free space, \(A\) is the area of the plates, and \(d\) is the separation between the plates.
The electric field \(E\) between the plates of the capacitor is given by:
\(E = \frac{V}{d}\)
Substituting \(V = Ed\) in the energy formula, we have:
\(U = \frac{1}{2}C(Ed)^2\)
Substituting the expression for \(C\), we get:
\(U = \frac{1}{2}\left(\frac{\epsilon_0 A}{d}\right)(Ed)^2\)
Upon simplifying, this expression becomes:
\(U = \frac{1}{2} \epsilon_0 E^2 A d\)
This matches the given option \(\frac{1}{2} \epsilon_0 E^2 A d\), confirming it as the correct choice.
Approach Solution -2
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: \[ \boxed{\frac{1}{2} \epsilon_0 E^2 A d}. \]
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