A parallel plate capacitor c0 is charged to Vo . The energy stored in the capacitor when the battery is disconnected and the plate separation is doubled is E1. The energy stored in the capacitor when the charging battery is kept connected and the separation bw the capacitor plates is doubled is E2. Then E1/E2 is

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Solution

$E 1 = \frac{1}{2} (\frac{ε_{0} A}{d}) V^{2} E_{2} = \frac{1}{2} (\frac{ε_{0} A}{2 d}) V^{2} \frac{E 1}{E 2} = 2 r a t i o i s 2 : 1 R e g a r d s$ Dear Student,

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