In the figure shown , after the switch 's' is turned from postion 'A' to postion 'B' the energy dissipated in the circuit in terms of capactance 'C' and total charge 'Q' is :

Options

A. `(1)/(8)(Q^(2))/(C)`
B. `(5)/(8)(Q^(2))/(C)`
C. `(3)/(4)(Q^(2))/(C)`
D. `(3)/(8)(Q^(2))/(C)`

Answer

D

Explanation

`q_(1)+q_(2)=CE implies q_(1)=(CE)/(4)implies q_(2)=(3CE)/(4)`
Initial energy `=(q_(1)^(2))/(2C)+(q_(2)^(2))/(2xx3C)=(C^(2)E^(2))/(32)+(9C^(2)E^(2))/(32xx3)=(4CE^(2))/(32),H=DeltaU=(CE^(2))/(2)-(CE^(2))/(8)=(3CE^(2))/(8)`