Question

A 4 micro farad capacitor charged by 200 volt supply. It is then dissconnected from the supply and is connected to anothe uncharged 2 micro farad capacitor. How much electrostatic energy of first capacitor is lost in the form of heat or e. m radiations?

Answer 1

Here,

C₁ = 4 μF = 4 ×10^-6F

V₁ = 200 volt

C₂ = 2 μF = 2 ×10^-6F

V₂ = 0

Initial Electrostatic energy stored in the capacitors will be

U_i = ½ C₁V₁² + ½ C₂V₂²

=>U_i = ½ ×4 ×10^-6 × (200)² + 0 = 8×10^-2J

After connecting the two capacitors, the total capacitance becomes

C = C₁ + C₂ = 4 ×10^-6 + 2 ×10^-6 = 6×10^-6 F

Charge in the first capacitor

Q = C₁V₁

=> Q = 4 ×10^-6×200 = 8×10^-4 C

Since, the 2^nd capacitor is uncharged so total charge on the system will be Q = 8×10^-4 C

Let V be the common potential of the capacitors after connection

V = Q/C

=> V = 8×10^-4/6×10^-6 = 4/3×10² volt

Now final electrostatic energy of the combination will be

U_f = ½CV² = ½Ã—6×10^-6×(4/3×10²)² = 4×10^-2 J = 5.33×10^-2 J

Now loss in energy in form of heat etc will be

U_i - U_f = (8-5.33) ×10^-2 J = 2.67 ×10^-2J