MHT CET · Physics · Capacitance
A parallel plate capacitor having plate area \(A\) and separation \(d\) is charged to a potential difference \(V\). The charging battery is disconnected and the plates are pulled apart to four times the initial separation. The work required to increase the distance between plates is:
- A \(\frac{\varepsilon_0 A V^2}{4 d}\)
- B \(\frac{2 \varepsilon_0 A V^2}{4 d}\)
- C \(\frac{\varepsilon_0 A V^2}{3 d}\)
- D \(\frac{3 \varepsilon_0 A V^2}{2 d}\)
Answer & Solution
Correct Answer
(D) \(\frac{3 \varepsilon_0 A V^2}{2 d}\)
Step-by-step Solution
Detailed explanation
Initial energy: \(\frac{q^2}{2 C}\)
The new capacitance is: \(C_n=\frac{C}{4}\)
Final energy after the change in capacitance is: \(\frac{4 q^2}{2 C}\)
Therefore, magnitude of work done is equal to the change in potential energy,
\(W=\frac{2 q^2}{C}-\frac{q^2}{2 C}=\frac{3 C V^2}{2}=\frac{3 \varepsilon_0 A V^2}{2 d}\)
The new capacitance is: \(C_n=\frac{C}{4}\)
Final energy after the change in capacitance is: \(\frac{4 q^2}{2 C}\)
Therefore, magnitude of work done is equal to the change in potential energy,
\(W=\frac{2 q^2}{C}-\frac{q^2}{2 C}=\frac{3 C V^2}{2}=\frac{3 \varepsilon_0 A V^2}{2 d}\)
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