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 capacitors. 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 = mu"F" = 4 xx 10^-6 "F"`

Supply voltage, V_{1} = 200 V

Electrostatic energy stored in C_{1 }is given by,

`"E"_1 = 1/2"C"_1"V"_1^2`

= `1/2 xx 4 xx 10^-6 xx (200)^2`

= `8 xx 10^-2 "J"`

Capacitance of an uncharged capacitor, `"C"_2 = 2mu"F" = 2 xx 10^-6 "F"`

When C_{2} is connected to the circuit, the potential acquired by it is V_{2}.

According to the conservation of charge, the 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"_1"V"_1`

`"V"_2 xx (4 + 2) xx 10^-6 = 4 xx 10^-6 xx 200`

`"V"_2 = 400/3` V

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)xx 10^-6 xx (400/3)^2`

= `5.33 xx 10^-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