A 700 pF capacitor is charged by a 50 V ...

A 700 pF capacitor is charged by a 50 V battery. The electrostatic energy stored it is

`6.7xx10^(-7)J`

`8.7xx10^(-7)J`

`9.7xx10^(7)J`

`6.7xx10^(7)J`

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The correct Answer is:

C = 700pF, V = 50V, E = ?
`E=1/2CV^(2)=(700xx10^(-12)xx25xx10^(2))/2`
`=8.7xx10^(-7)J`

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