Question

Derive expression for energy stored in capacitor show that whatever 2 conductors sahre charges by brinhing them into electrical contact there is a loss of energy

Answer 1

Both conductors have different potentials initially.

Considering the energy of the system (including both conductors)

E₁= (1/2)C₁V₁² + (1/2)C₂V₂²----- (1)

Where C₁ and C₂ are capacitance of the two capacitors and V₁ and V₂ be the potentials of the two capacitors

When the two capacitors are brought in electrical contact, potential difference for both will be common.

The common potential V= (C₁V₁+C₂V₂)/(C₁+C₂)----- (2)

So, energy of the system is

E₂= (1/2)(C₁+C₂)V²      

From (2), E₂= (1/2)(C₁+C₂)(C₁V₁+C₂V₂)²/(C₁+C₂)²

                  = (1/2){(C₁V₁+C₂V₂)²/(C₁+C₂)}----- (3)

Now, E₁-E₂= [(1/2)C₁V₁² + (1/2)C₂V₂²]- [(1/2)(C₁V₁+C₂V₂)²/(C₁+C₂)]

                 = C₁C₂(V₁-V₂)²/2(C₁+C₂)