AP EAMCET · PHYSICS · Capacitance
The energy of a parallel plate capacitor when connected to a battery is \(E\). With the battery still in connection, if the plates of the capacitor are separated, so that the distance between them is twice the original distance, then the electrostatic energy becomes
- A \(2\ E\)
- B \(\frac{E}{4}\)
- C \(\frac{E}{2}\)
- D \(4\ E\)
Answer & Solution
Correct Answer
(C) \(\frac{E}{2}\)
Step-by-step Solution
Detailed explanation
\begin{aligned} & \text { } E=\frac{1}{2} C V^2 \Rightarrow C \propto \frac{1}{\mathrm{~d}} \\ & \Rightarrow \frac{C_1}{C_2}=\frac{d_1}{d_2}=\frac{2 d}{d}=2\left[\because C=\frac{A \in_0}{d}\right] \\ & \text { Now, } \frac{E_1}{E_2}=\frac{C_1}{C_2}=\frac{C}{C /…
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