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 capacitors. How much electrostatic energy of the first capacitor is lost in the form of heat and electromagnetic radiation?

Answer 1

Capacitance of a charged capacitor, `"C"_1 = mu"F" = 4 xx 10^-6  "F"`

Supply voltage, V₁ = 200 V

Electrostatic energy stored in C₁ 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 C₂ is connected to the circuit, the potential acquired by it is V₂.

According to the conservation of charge, the initial charge on capacitor C₁ is equal to the final charge on capacitors, C₁ and C₂.

∴ `"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 C₁

= E₁ − E₂

= 0.08 − 0.0533

= 0.0267

= 2.67 × 10⁻² J