The switch shown in figure is suddenly closed. Find the current through the resistor when energy dissipated in the resistor is one third of total energy dissipated in the resistor.

Answer

`I=sqrt((2)/(3)). (Q)/(4pi in_(0)R R_(0))`

Explanation

Initial energy stored in conducting sphere
`U = (Q^(2))/(2C)=(Q^(2))/(8H in_(0)R)`…..(i)
when switch is closed, all energy dissipated in the resistor, according to question.
Let 'q' charge on the sphere when one third energy is lost
`:. (q^(2))/(2c)=(2u)/(3),(q^(2))/(8pi(in_(0)R))=(2)/(3)(Q^(2))/(8 pi (in_(0)R))`
`:. q=sqrt((2)/(3))Q`....(ii)
Hence point of sphere at the instant
= point dift. across resister at that instant
`:. (q)/(4pi(in_(0)R))=I.R_(0)`
`:. I =sqrt((2)/(3)).(Q)/(4 pi in_(0)R R_(0))`