A 600pF capacitor is charged a 200V supply. It is then disconnnected from the supply and is connected to another uncharged 600pF capacitor. How much electro static energy is lost in the process.

Explanation

`C_(1)=600pF, V_(1)=200V," "C_(2)=600pF, triangleE=?`
`E_(1)=1/2 C_(1)V_(1)^(2)=1/2 xx 600 xx 10^(-12) xx 200 xx 200=12 xx 10^(-6)J`
`E_(2)=1/2 (Q^(2))/(C_(1)+C_(2))=1/2 xx ((C_(1)V_(1))^(2))/(C_(1)+C_(2))=1/2 xx ((600 xx 10^(-12) xx 200)^(2))/(1200 xx 10^(-12))`
`=1/2 xx ((600)^(2) xx (10^(-2))^(2) xx 200 xx 200)/(1200 xx 10^(-12))=(600 xx 600 xx 10^(-12) xx 100)/(6)=6 xx 10^(-6)J`
`triangleE=6 xx 10^(-6)J`