Find the energy stored in a capacitor of capacitance `100muF ` when it is charged to a potential difference of 20 V.

Explanation

The energy stored in the capacitor is `
`U = (1)/(2) CV^2 = (1)/(2) (100 mu F) (20 V)^2 = 0.02 J`
`. The energy stored in a capacitor is electrostatic potential energy. When we pull the plates of a capacitor apart, we have to do work against the electrostatic attraction between the plates. In which region of space is the energy stored ? When we increase the separation between the plates from `
`d_1 to d_2`
`, an amount`
`(Q)^2/(2A epsilon_0) (d_2 - d_1)`
` of work is performed by us and The field due to the charges `
`Q_p, - Q_p`
` is directed oppositely and has magnitude `
`E_p =(sigma_p )/(epsilon_o) =(Q_p)/(Aepsilon_o)`
`. The resultant field is `
` E = E_0 - E_p`
` = (Q-Q_p)/(A epsilon_0)`
`. From equations (ii) and (iii), (Q - Q_p)/(epsilon_o A) = (Q)/(epsilon _o AK)`
` or, `
`Q - Q_p = (O)/(K)`
` or, `
`Q_p = Q (1 -(1)/(K)).