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`

= `1/2 xx (600 xx 10^-12) xx (200)^2`

= `1.2 xx 10^-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 = 2/600 = 1/300`

C' = 300 pF

New electrostatic energy can be calculated as

`"E'" = 1/2 xx "C'" xx "V"^2`

= `1/2 xx 300 xx (200)^2`

= `0.6 xx 10^-5 "J"`

Loss in electrostatic energy = E − E'

= `1.2 xx 10^-5 - 0.6 xx 10^-5`

= `0.6 xx 10^-5`

= `6 xx 10^-6 "J"`

Therefore, the electrostatic energy lost in the process is `6 xx 10^-6 "J"`.