A 4 `mu`F capacitor is charged by a 200 V supply. It is then disconnected from the supply, and is connecled to another uncharged 2 `mu`f capacitor. How much electrostatic energy of the first capacilor is lost in the form of heat and electromagnetic radiation ?

Explanation

Energy stored in capacitor of 4 `mu`F initially .
`U_(i)=(1)/(2) CV^(2)= (1)/(2) xx4xx10^(-6)xx(200)^(2)`
`U_(i) = 8xx10^(-2) J`
Charge on capacitor of `4 muF`
`Q =CV = 4xx10^(-6) xx200`
`:. Q = 8xx10^(-4)` C
Potential of combination of capacitors `4muF` and `2 muF`
`V_(1) = ("Total charge")/("Total capacitance") = (0+8xx10^(-4))/((4+2)xx10^(-6))`
`:.V_(1)=(8xx10^(-4))/(6xx10^(-6))=(800)/(6) =(400)/(3) V `
Final electrostatic energy of combination
`U_(f) = (1)/(2) (C_(1)+C_(2))V_(1)^(2)`
`=(1)/(2)(4+2)xx10^(-6)xx((400)/(3))^(2)`
`=(1)/(2)xx(6xx10^(-6)xx16xx10^(4))/(9)`
`:. U_(f) =(16)/(3) xx10^(-2)J`
`= 5.33xx1^(-2) J`
Hence amount of electrostatic energy lost by capacitor of `4 muF `
`DeltaU = U_(i) -U_(f)`
`= (8-5.33) xx10^(-2) J`
`= 2.67xx10^(-2) J`