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
- A \(\dfrac{1}{2}\varepsilon_0 E^2\)
- B \(\varepsilon_0 E a d\)
- C \(\dfrac{1}{2}\varepsilon_0 E^2 A d\)
- D \(\dfrac{E^2 A d}{\varepsilon_0}\)
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: \(\dfrac{1}{2}\varepsilon_0 E^2 A d\)
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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