Energy loss in redistribution of charge:

#Energy loss in redistribution of charge:

Let two conductor A and B of capacities C1 and C2. on charging them by charge Q1 and Q2 respectively then their potential become V1 and V2 here V1>V2 .

Thus before joining:
charge on conductor A, Q1 = C1.V1
Charge on conductor B, Q2 = C2.V2

Electric energy stored in conductor A is
U1 = 1/2 C1.V1²

Electric energy stored in conductor B is
U2 = 1/2 C2.V2²

Hence, Total energy before joining is
U = U1 + U2
U = 1/2 C1.V1² + 1/2 C2.V2²
U = 1/2(C1.V1² + C2.V2²)

After joining :
Two charged conductor joining together by the conducting wire L then charge flows from high potential to low potential till both the conductor getting common potential V.
In this process electrical energy is converted into heat energy due to resistance of conducting wire. This heat energy expelled in the atmosphere which is known as loss of energy.

Let the common potential be V
Both the conductor after joining behave like a single conductor.

Thus, total charge,.  Q = Q1 + Q2
                            Q = C1.V1 + C2.V2


Total capacity, C = C1 + C2

By the definition of capacitance
C = Q/V
Hence, V = Q/C

Putting the values

[V= (C1.V1+C2.V2) / (C1+C2) ]

This is common potential after joining

Total Electric energy of conductor after joining     U'  = 1/2 C.V²

U' = 1/2(C1+C2) [(C1.V1+C2.V2) / (C1+C2)]²

U' =1/2(C1+C2) × (C1².V1²+C2².V2²+2C1.V1×C2.V2) /(C1+C2)²

U' = 1/2(C1².V1²+C2².V2²+2C1.V1×C2.V2) /                                                       (C1+C2)

Thus loss of energy, ∆U = U - U'

∆U = 1/2(C1.V1² + C2.V2²) --
1/2(C1².V1²+C2².V2²+2C1.V1×C2.V2)/(C1+C2)

On solving above equation which is so long we get the given result

{∆U = 1/2 [C1.C2( V1 - V2)²]/(C1+C2) }