AP EAMCET · PHYSICS · Capacitance
Between the plates of a parallel plate capacitor of plate area \(\mathrm{A}\) and capacity \(0.025 \mu \mathrm{F}\), a metal plate of area, \(\mathrm{A}\) and thickness equal to \(\frac{1}{3}\) of the separation between the plates of the capacitor is introduced. If the capacitor is charged to \(100 \mathrm{~V}\), then the amount of work done to remove the metal plate from the capacitor is
- A \(62.5 \mu \mathrm{J}\)
- B \(30.2 \mu \mathrm{J}\)
- C \(52.6 \mu \mathrm{J}\)
- D \(35.4 \mu \mathrm{J}\)
Answer & Solution
Correct Answer
(A) \(62.5 \mu \mathrm{J}\)
Step-by-step Solution
Detailed explanation
Initially \(c^{\prime}=\frac{\varepsilon A}{d-\frac{d}{3}}=\frac{3}{2} \frac{\varepsilon_0 A}{d}=\frac{3}{2} c\) [C=capacitor of plate without conduction] \(\begin{aligned} & v_i=v_f=100 v \\ & u_i=\frac{1}{2} c v^2 \\ & u_f=\frac{1}{2} c v^2\end{aligned}\)…
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