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"`.
