Question

how is energy lost due to sharing of charge between the conductors

Answer 1

There is actually no major loss of energy due to sharing of charge between two capacitors chosen as conductors. However, some energy is generated in the form of heat while some energy gets scattered. When the two capacitors are connected in parallel, charge will flowbetween the capacitors in both directions.

Answer 2

When two conductors of different potentials are brought close to each other, charge is shared between them. During this process, some amount of energy is lost which is disappeared in the form of heat.
In order to calculate the total loss of energy during this process, consider two capacitors having capacitance C₁ and C₂ having distinct potentials V₁ and V₂ respectively. Now we know that, the charge always flow from a capacitor of high potential to a capacitor of low potential.
Hence,
Their common potential is given as:
V = (C₁V₁ + C₂ V₂)  / C₁ + C₂  .......... (1)
Total energy of the system will be:
E₁ = 1/2C₁V₁² + 1/2 C₂ V₂²   ........... (2)
E₂ = 1/2 (C₁ + C₂) V²
E₂ = [ 1/2 (C₁ + C₂) ] . [ (C₁V₁ + C₂ V₂)  / (C₁ + C₂) ]²  (Using equation (1))
E₂ = [ 1/2 (C₁ + C₂) ] . [ (C₁V₁ + C₂ V₂)² / (C₁ + C₂)² ]
E₂ = (C₁V₁ + C₂ V₂)² / 2 (C₁ + C₂)   .......... (3)
Subtracting equation (2) and equation (3), we get:
E₁ - E₂ = [1/2 C₁V₁² + 1/2 C₂ V₂² ] - [ (C₁V₁ + C₂ V₂)² / 2 (C₁ + C₂) ]
E₁ - E₂ = [ C₁V₁² (C₁ + C₂) + C₂ V₂² (C₁ + C₂) - (C₁V₁ + C₂ V₂)² ] / 2 (C₁ + C₂)
E₁ - E₂ =C₁²V₁²+C₁C₂V₁²+C₁C₂V₂²+C₂²V₂²-(C₁²V₁²+C₂²V₂² +2C₁C₂V₁V₂)/2(C₁+C₂)
E₁ - E₂ =C₁²V₁²+C₁C₂V₁²+C₁C₂V₂²+C₂²V₂²- C₁²V₁²-C₂²V₂² - 2C₁C₂V₁V₂ /2(C₁ + C₂)
E₁ - E₂ = C₁C₂ (V₁ - V₂)² / 2(C₁ + C₂)
The above equation shows that E₁ - E₂ is greater then zero.
Or E₁ is greater then E₂
Or E₂ is less then E₁
Hope it helps. Thanks.