MHT CET · Physics · Capacitance
The energy required to charge a parallel plate condenser of plate separation \(d\) and plate area of cross-section \(A\) such that the uniform electric field between the plates is \(E\), is
- A \(\frac{1}{2} \varepsilon_{0} E^{2} / A d\)
- B \(\varepsilon_{0} E^{2} / A d\)
- C \(\varepsilon_{0} E^{2} A d\)
- D \(\frac{1}{2} \varepsilon_{0} E^{2} A d\)
Answer & Solution
Correct Answer
(C) \(\varepsilon_{0} E^{2} A d\)
Step-by-step Solution
Detailed explanation
Energy given by the cell
\(E=C V^{2} \)
\( \text {Here, } C =\text { capacitance of condenser } \)
\( =\frac{A \varepsilon_{0}}{d}\)
\(V=\) potential difference across the plates \(=E d\)
\(
\text {Therefore, } E=\left(\frac{A \varepsilon_{0}}{d}\right)(E d)^{2}=A \varepsilon_{0} E^{2} d
\)
\(E=C V^{2} \)
\( \text {Here, } C =\text { capacitance of condenser } \)
\( =\frac{A \varepsilon_{0}}{d}\)
\(V=\) potential difference across the plates \(=E d\)
\(
\text {Therefore, } E=\left(\frac{A \varepsilon_{0}}{d}\right)(E d)^{2}=A \varepsilon_{0} E^{2} d
\)
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