Question

(a) Derive an expression for common potential when two capacitors at different potentials touch each other.

(b) Two capacitors at different potentials touch each other. Show that there is always loss of energy.

Answer 1

(a) Let two capacitors of capacities C₁, C₂ and potentials V₁ and V₂ touch each other. If their common potential is V, then the sum of charges

Before sharing = C₁V₁ + C₂V₂

And after sharing = C₁V + C₂V

= V (C₁ + C₂)

Since charges are conserved, therefore

V(C₁ + C₂) = C₁V₁ + C₂V₂

or V = \(\frac{C_1V_1 + C_2V_2}{C_1 + C_2}\)

(b) Sum of energies before sharing is

E₁ = \(\frac{1}{2}C_1V_1^2 + \frac{1}{2}C_2V_2^2\)

and sum of energies after sharing is

E₁ - E₂ = \(\frac{C_1C_2(V_1^2+V_2^2-2V_1V_2)}{2(C_1 +C_2)}\)

= \(\frac{C_1C_2(V_1-V_2)^2}{2(C_1 +C_2)}\)

= a positive quantity.

Hence there is always a loss of energy when two capcaitors share the charges.