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

A 600 pF capacitor is charged by a 200V supply. It is then discoconnected 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 = 600pF " and " V=200V`.
`:. E = (1)/(2)CV^2 = (1)/(2)xx600xx10^(-12)xx(200)^2= 12xx10^(-6)J`
When this capacitors is connected with another capacitor of same capacity, the potential will be shared equally and the capacitance will be added.
`:. V' = (V)/(2)= 100V " and "C' = 2C = 1200xx10^(-12)F`
`E' = (1)/(2)C'V'^2 = (1)/(2)xx1200xx10^(-12)xx(100)^2 = 6xx10^(-6) J`
Loss of electrostatic energy
`= E - E' = (12-6)xx 10^(-6) J = 6xx10^(-6)J`.
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