Two identical thin rings each of radius R are coaxially placed at distance R apart . If `Q_(1) and Q_(2)` are the charges uniformly spread on the two rings , the work done is moving a charge from centre of one ring to the order is :

Options

A. zero
B. `(q(Q_(1)-Q_(2))(sqrt(2)-1))/(sqrt(2).4piepsilon_(0)R)`
C. `(qsqrt(2)(Q_(1)+Q_(2)))/(4piepsilon_(0)R)`
D. `(q((Q_(1))/(Q_(2)))(sqrt(2)+1))/(sqrt(2).4epsilon_(0)R)`

Answer

B

Explanation

`W=q (V_(A)-V_(B))`

`V_(A) = 1/(4piepsilon_(0)) [ (Q_(1))/R +(Q_(2))/(sqrt(2R))] `
`V_(B) = 1/(4piepsilon_(0)) [ (Q_(2))/R + Q_(1)/(sqrt(2R))] `
Then `W = (q(Q_(1)-Q_(2)))/(sqrt(2) . 4piepsilon_(0)R) . (sqrt(2)-1)` .