A parallel plate capacitor of capacity C0 is charged to a potential V0.1) The energy stored in the capacitor when the battery is disconnected and the plate separation is doubled is E1.2) The energy stored in the capacitor when the charging battery is kept connected and the separation between the capacitor plates is doubled is E2. Then E1/E2

When battery is disconnected, Uo=Q22cQ is same When battery is kept connected.U0=12CV2V is same E1=Q22×C02=Q2C0=C0V02E2=12C02V02=C0V024⇒E1E2=41

A parallel plate capacitor of capacity C0 is charged to a potential V0.1) The energy stored in the capacitor when the battery is disconnected and the plate separation is doubled is E1.2) The energy stored in the capacitor when the charging battery is kept connected and the separation between the capacitor plates is doubled is E2. Then E1/E2

When battery is disconnected, Uo=Q22cQ is same When battery is kept connected.U0=12CV2V is same E1=Q22×C02=Q2C0=C0V02E2=12C02V02=C0V024⇒E1E2=41

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A parallel plate capacitor of capacity C₀ is charged to a potential V₀.
1) The energy stored in the capacitor when the battery is disconnected and the plate separation is doubled is E₁.
2) The energy stored in the capacitor when the charging battery is kept connected and the separation between the capacitor plates is doubled is E₂. Then E₁/E₂

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Detailed Solution

When battery is disconnected, $U_{o} = \frac{Q^{2}}{2 c}$
Q is same When battery is kept connected.
$U_{0} = \frac{1}{2} {CV}^{2}$
V is same $E_{1} = \frac{Q^{2}}{2 \times \frac{C_{0}}{2}} = \frac{Q^{2}}{C_{0}} = C_{0} V_{0}^{2}$
$E_{2} = \frac{1}{2} \frac{C_{0}}{2} V_{0}^{2} = \frac{C_{0} V_{0}^{2}}{4} \Rightarrow \frac{E_{1}}{E_{2}} = \frac{4}{1}$