#### Question

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

#### Solution

Capacitance of a charged capacitor, `C_1=muF=4xx10^-6 F`

Supply voltage, *V*_{1} = 200 V

Electrostatic energy stored in *C*_{1 }is given by,

`E_1=1/2C_1V_1^2`

`=1/2xx4xx10^-6xx(200)^2`

`=8xx10^-2 J`

Capacitance of an uncharged capacitor, `C_2=2muF=2xx10^-6F`

When *C*_{2} is connected to the circuit, the potential acquired by it is *V*_{2}.

According to the conservation of charge, initial charge on capacitor *C*_{1} is equal to the final charge on capacitors, *C*_{1} and *C*_{2}.

`V_2(C_1+C_2)=C_1V_1`

`V_2xx(4+2)xx10^-6=4xx10^-6xx200`

`V_2=400/3V`

Electrostatic energy for the combination of two capacitors is given by,

`E_2=1/2(C_1+C_2)V_2^2`

`=1/2(2+4)xx10^-6xx(400/3)^2`

`=5.33 xx10^-2 `J

Hence, amount of electrostatic energy lost by capacitor *C*_{1}

= *E*_{1} − *E*_{2}

= 0.08 − 0.0533 = 0.0267

= 2.67 × 10^{−2} J