A parallel-plate capacitor of capacity C0 is charged to a potential V0. (i) The energy stored in the capacitor when the battery is disconnected and the plat separation is doubled is E1. (ii) The energy stored in the capacitor when the charging battery is kept connected and the separation between the capacitor plates is doubled is E2. Find the ratio E1/E2.

Initial energy: E=12C0V02=12Q02C0i) E1 = 2E, because charge remains the same and the capacitance becomes half. ii) E2 =E/2, because potential remains the same and the capacitance becomes half.E1E2=4

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A parallel-plate capacitor of capacity C₀ is charged to a potential V₀. (i) The energy stored in the capacitor when the battery is disconnected and the plat separation is doubled is E₁. (ii) The energy stored in the capacitor when the charging battery is kept connected and the separation between the capacitor plates is doubled is E₂. Find the ratio E₁/E₂.

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Detailed Solution

Initial energy: $E = \frac{1}{2} C_{0} V_{0}^{2} = \frac{1}{2} \frac{{Q_{0}}^{2}}{C_{0}}$

i) E₁ = 2E, because charge remains the same and the capacitance becomes half.

ii) E₂ =E/2, because potential remains the same and the capacitance becomes half.

$\frac{E_{1}}{E_{2}} = 4$

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