Two charge + Q each are at a distance r from each other. A third charge q is placed in the line joining above two charges, so that all of them are in equilibrium. Find value of q.

Options

A. `Q/2`
B. `Q/3`
C. `Q/4`
D. `Q/6`

Answer

C

Explanation

`1/(4piepsilon_(0)) . (Q.q)/(x^(2)) = 1/(4piepsilon_(0)) . (Qq)/((r-x)^(2))`
` :. " " x = r - x , x = r//2 `
Again for + Q on left ,
`1/(4piepsilon_(0)) (Q.q)/(x^(2)) = 1/(4piepsilon_(0)) (Q.Q)/(r^(2))`

` :. " " q/(x^(2)) = Q/(r^(2) ` , But x = r/2
`(4q)/(r^(2)) = Q/(r^(2)) , q = Q/4` .