AP EAMCET · PHYSICS · Capacitance
A parallel plate capacitor having capacity \(\mathrm{C}_0\) is charged to \(\mathrm{V}_0\). With battery disconnected, if the separation between the plates is doubled then the energy stored in it is \(E_1\). Instead if the separation between the plates is doubled, with battery in connection, the energy stored in it is \(\mathrm{E}_2\). Then the value of \(\frac{E_2}{E_1}\) is
- A \(0.5\)
- B \(1.5\)
- C \(2\)
- D \(0.25\)
Answer & Solution
Correct Answer
(D) \(0.25\)
Step-by-step Solution
Detailed explanation
When battery disconnected, charge is constant…
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