A parallel plate capacitor of capacity {C_0} is charged to a potential {V_0} (i) energy stored in ca

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2. Electric Potential and Capacitance

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A parallel plate capacitor of capacity ${C_0}$ is charged to a potential ${V_0}$

$(i)$ The energy stored in the capacitor when the battery is disconnected and the separation is doubled ${E_1}$

$(ii)$ The energy stored in the capacitor when the charging battery is kept connected and the separation between the capacitor plates is doubled is ${E_2}.$

Then ${E_1}/{E_2}$ value is

A

$4$

B

$1.5$

C

$2$

D

$0.25$

Solution

(a) Let $E = \frac{1}{2}{C_0}{V_0}^2\,{\rm{then}}\,{\rm{ }}{E_{\rm{1}}} = 2E$ and ${E_2} = \frac{E}{2}$
So $\frac{{{E_1}}}{{{E_2}}} = \frac{4}{1}$

Standard 12

Physics

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