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

Explanation

Capacitance of first capacitor
`C_1=4xx10^-6F`
`V_1=200V`
Initial electrostatic energy id `C_1` is
`E_1=1/2C_1V_1^2`
`=1/2xx4xx10^-6(200)^2`
`E_1=8xx10^-2J` …...(1)
Capacitance of another capacitor
`C_2=4muF`
When `C_2` is connected to `C_1`, charge flows and both acquire a common potential
`V = "total charge"/"total capacitance"`
`V= (C_1V_1)/(C_1+C_2)`
`=(4xx10^-6xx200)/((4+2)xx10^-6)=800/6V`
Final electrostatic energy of both capacitors is
`E_2=1/2(C_1+C_2)V^2`
`E_2=1/2xx6xx10^-6xx800/6xx800/6`
`=5.33xx10^-6J` .....(2)
From equation (1) and () we get energy dissiputed in the form of heat
`=E_1-E_2=8xx10^-2-5.33xx10^-2`
`=2.67xx10^-2J`