(a) A 900 pF capacitor is charged by 100 V battery [Fig. 2.31(a)]. How much electrostatic energy is stored by the capacitor?
(b) The capacitor is disconnected from the battery and connected to another 900 pF capacitor [Fig. 2.31(b)]. What is the electrostatic energy stored by the system?

Explanation

(A) Here C = 900 p F = `9xx10^(-10)` F
V= 100 V
`:.` Charge on capacitor Q = CV
`:. Q = 9xx10^(-10) xx100`
`:. Q = 9xx10^(-8) C `
The energy stored by the capacitor is
`U =(1)/(2) CV^(2) = (1)/(2)xxQV [ becuse ` CV = Q]
`= (1)/(2)xx9xx10^(-8)xx100`
` = 4.5 xx10^(-6) ` J
(b) Now given capacitor is removed from battery and connect with similar type chargeless capacitor then the charge on both capacitor
`Q^(1) = (9xx10^(-8))/(2) = 4.5xx10^(-8)` C
In this situation the potential energy of the system ,
`U=(1)/(2) ((Q^(1))^(2))/(C) =((4.5xx10^(-8))^(2))/(2xx9xx10^(-10))`
`:. U = 2.25xx10^(-6)` J