Two point charges `+4q` and `+q` are placed at a distance `L` apart. `A` third charge `Q` is so placed that all the three charges are in equilibrium. Then location. And magnitude of the third charge will be

Options

A. `Q = (4q)/(9) , x = (d)/(3)`
B. `Q = (4q)/(9) , x = (d)/(4)`
C. `Q = -(4q)/(9) , x = (d)/(3)`
D. `Q = -(4q)/(9) , x = (d)/(4)`

Answer

3

Explanation

The system is in equilibrium i.e. net electric force on each charge is zero.
At `Q`: Force due to `q,F_(1) = (1)/(4 pi in_(0)) (qQ)/(x^(2))` , along `AB`
Force due to `4q , F_(2) = (1)/(4 pi in_(0)) (4qQ)/((d - x)^(2))` along `BA`
Since net force on `Q` is zero ,
`F_(1) = F_(2) rArr (1)/(x^(2)) = (4)/((d - x)^(2))`
`(1)/(x) = (2)/((d - x)) rArr d - x = 2x rArr x = (d)/(3)`
`A + q` : force due to `Q , F_(1) = (1)/(4 pi in_(0)) (qQ)/(x^(2))` ,along `BA`
Force due to `4q , F_(2) = (1)/(4 pi in_(0)) (q.4q)/(d^(2))` , along `BA`
Since net force on `q` is zero
`F_(1) + F_(2) = 0 rArr (qQ)/(x^(2)) + (4q^(2))/(d^(2)) = 0`
`(Q)/((d //3)^(2)) + (4q)/(d^(2)) = 0 rArr 9 Q + 4Q = 0`
`Q = -(4q)/(9)`