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. What is the common potential in `V` and energy lost in `J` afrte reconnection?

Explanation

Capacitance of the capacitor, C=600 pF
potential diffference V=200 V
Elec trostatic energy stored in the capacitor is given by,
`E=1/2 CV^(2)`
`=1/2 xx(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=2/600 =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)-0.16xx10^(-5)`
`=0.6xx10^(-5)`
`=6xx10^(-6) J`
Therefore, the electrostatic energy lost in the process is `6xx10^(-6)` J