Two point charges `q_(1) = +2 C` and `q_(2) = - 1C` are separated by a distance `d`. The position on the line joining the two charges where a third charge `q = + 1 C` will be in equilibrium is at a distance

Options

A. `(d)/(sqrt(2))` from `q_(1)` and `q_(2)`
B. `(d)/(sqrt(2))` from `q_(1)` away from `q_(2)`
C. `(d)/(sqrt(2) - 1)` from `q_(2)` between `q_(1)` and `q_(2)`
D. `(d)/(sqrt(2) - 1)` from `q_(2)` away from `q_(1)`

Answer

4

Explanation

Net force on `q` will not be zero if it is placed between charges. The third charge should be placed nearer the smaller charge outside `AB`.Let `q` is placed at distance `x` from `qq_(2) = -1 C`.
`(q_(1) ,q) : F_(1) = (1)/(4 pi in_(0)) (q_(1) q)/((d + x)^(2))` along `AP`
`(q_(2) ,q) : F_(2) = (1)/(4 pi in_(0)) (q_(2) q)/(x^(2))` along `PB`
`F_(1) = F_(2)`
`(q_(1) q)/((d + x)^(2)) = (q_(2) q)/(x^(2))`
`(2)/((d + x)^(2)) = (1)/(x^(2)) rArr (sqrt(2))/(d + x) = (1)/(x)`
`(sqrt(2) - 1) = d rArr x = (d)/((sqrt(2) - 1))`