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A 600 pF capacitor is charged by a 200 V...

A 600 pF capacitor is charged by a 200 V supply. It is then disconnected from the supply and is connected to another uncharged 600 pF capacitor. How much electrostatic energy is lost in the process ?

Text Solution

Verified by Experts

Here `C_1 = C_2 = 600 pF = 6 xx 10^(-10) F, V_1 = 200 V and V_2 = 0`
Loss of electrostatic energy on sharing charges
`Delta u = (C_1C_2(V_1-V_2)^2)/(2(C_1 + C_2)) = (6 xx 10^(-10) xx 6 xx 10^(-10) xx (200-0)^2)/(2 (6 xx 10^(-10) + 6 xx 10^(-10))`
`=(6 xx 10^(-10) xx 6 xx 10^(-10) xx 200 xx 200)/(2 xx 12 xx 10^(-10)) =6 xx 10^(-6)J`.
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