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:

Answer

D

Explanation

Let q charge is situated at themidposition of the lineAB.The distance between AB is x. A and B be the positions of charges Q and Q respectively.

Let `AC = (x)/(2), BC = (x)/(2)`
The force on A due to charge q at C.
`vecF_(CA) = (1)/(4piepsi_(0)). (Q.q)/((x//2)) ` a long `vecAC`
The system is in equilibrium, then two oppositely directed force must be equal,i.e., total force on A is equal to zero.
`vecF _(CA) + vecF_(AB) = 0 rArr vecF_(CA) = - vecF_(AB)`
` (1)/(4piepsi_(0)).(4Q_(q))/(x^(2)) = (-1)/(4piepsi_(0)) , (Q^(2))/(x^(2))`
`rArr q = - (Q)/(4)`