A 600 pF capacitor is charged by a 200 V supply. It is then disconnected from the supply and is connected to another uncharged 600 pF capacitor. How much electrostatic energy is lost in the process?

#### Solution

Capacitance of the capacitor, *C* = 600 pF

Potential difference, *V* = 200 V

Electrostatic energy stored in the capacitor is given by,

`E=1/2 CV^2`

`=1/2xx(600xx10^-12)xx(200)^2`

`=1.2xx10^-5 J`

If supply is disconnected from the capacitor and another capacitor of capacitance *C* = 600 pF is connected to it, then equivalent capacitance (*C*^{’}) of the combination is given by,

`1/C'=1/C+1/C`

`=1/600+1/600=1/300`

`C'=300 pF`

New electrostatic energy can be calculated as

`E'=1/2xxC'xxV^2`

`=1/2xx300xx(200)^2`

`=0.6xx10^-5 J`

Loss in electrostatic energy=E-E'

`=1.2xx10^-5-0.6xx10^-5`

`=0.6xx10^-5`

`=6xx10^-6 J`

Therefore, the electrostatic energy lost in the process is `6xx10^-6 J`.