Two identical thin rings, each of radius R, are coaxially placed at a distance R. If `Q_(1) and Q_(2)` are respectively, the charges uniformly spread on the two rings, find the work done in moving a charge q from centre of ring having charge `Q_(1)` to the other ring.

Answer

`q(Q_1 - Q_2)(sqrt(2 )- 1)//sqrt(2) R . 4 pi epsilon_0`

Explanation

Work done = charge xx difference in potential Let the particle move from O to O' (Fig. S3.19) Potential at point O is
`V = (1)/(4 pi epsilon_0) Q_(1)/(R)+ Q_(2)/(sqrt(2) R)`
Potential at point O' is
`V' = (1)/(4 pi epsilon_0)[Q_(2)/R + Q_(1)/(sqrt(2) R)]`

Difference in potential is
`Delta V = V - V'`
=`(1)/(4 pi epsilon_0) [Q_(1)/R + Q_(2)/(sqrt(2) R)] - (1)/(4 pi epsilon_0) [Q_(2)/R + Q_(1)/(sqrt(2 )R)`
=`(1)/(4 pi epsilon_0) [(1)/R (Q_1 - Q_2) + (1)/(sqrt(2) R) - (Q_2 - Q_1)]`
=`(1)/(4 pi epsilon_0) [(Q_1 - Q_2)/R (1 - (1)/sqrt(2))]`
`:. W d = q Delta V = q(1)/(4 pi epsilon_0) [((Q_1 - Q_2))/(sqrt(2) R) (sqrt(2 - 1))]`.