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

Explanation

Capacitance of a charged capacitor , `C_(1)=4muF=4xx10^(-6) F`
supply voltage, `V_(1)=200V`
electrostatic energy stored in `C_(1)` in given by
`E_(1)=1/2 C_(1)V_(1)^(2)`
`=1/2xx4xx10^(-6)xx(200)^(2)`
`=8xx10^(-2)J`
Capacitance of an uncharged capacitor, `C_(2)=2muF=2xx10^(-6)F`
when `C_(2)` is connected to the circuit, the potential acquired by it is `V_(2)`
According to the conservation of charge, intial charge capacitor `C_(1)` is equal to the final charge on capacitors `C_(1)` and `C_(2)`
`:. V_(2)(C_(1)+C_(2))=C_(1)V_(1)`
`V_(2)xx(4+2)xx10^(-6)=4xx10^(-6)xx200`
`V_(2)=400/3 V`
Electrostatic energy for the combination of two capacitors is given by,
`E_(2)=1/2(C_(1)+C_(2))V_(2)^(2)`
`=1/2(2+4)xx10^(-6)xx(400/3)^(2)`
`=5.33xx10^(2)J`
Hence , amount of electrostatic energy lost by capacitor `C_(1)`
`=E_(1)-E_(2)`
`=0.0800.0533 =0.0267`
`=2.67xx10^(-2)J`