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KCET · Physics · Capacitance

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 energy stored in the capacitor is

  1. A \(\dfrac{1}{2}\varepsilon_0 E^2\)
  2. B \(\varepsilon_0 E a d\)
  3. C \(\dfrac{1}{2}\varepsilon_0 E^2 A d\)
  4. D \(\dfrac{E^2 A d}{\varepsilon_0}\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(\dfrac{1}{2}\varepsilon_0 E^2 A d\)

Step-by-step Solution

Detailed explanation

Capacitance of a parallel plate capacitor is given by \(C = \dfrac{\varepsilon_0 A}{d}\)

Potential difference between the plates is \(V = E d\)

Energy stored in the capacitor is \(U = \dfrac{1}{2} C V^2\)

Substituting the values of \(C\) and \(V\):

\(U = \dfrac{1}{2} \left( \dfrac{\varepsilon_0 A}{d} \right) (E d)^2\)

\(U = \dfrac{1}{2} \varepsilon_0 E^2 A d\)

Answer: \(\dfrac{1}{2}\varepsilon_0 E^2 A d\)
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