A charge q is placed at the centre of the line joining two equal charges Q. The system of the three charges will be in equilibrium if q is equal to:

Options

A. `-(q)/(2)`
B. `-(q)/(4)`
C. `+(q)/(4)`
D. `+(q)/(2)`

Answer

2

Explanation

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