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

Options

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

Answer

D

Explanation

Charge q has to be negative
Repulsive forces on charge +Q (at A) due to charge
+Q (at B)
`F=(1)/(4piepsi_(0))*(QxxQ)/(a^(2))`
Force of attraction on charge +Q (at A) due to charge -q (at C)
`F.=(1)/(4piepsi_(0))*(Qxxq)/((a//2)^(2))`
For equilibrium of charge +Q (at A), `F=F.`
`(1)/(4piepsi_(0))*(QxxQ)/(a^(2))=(1)/(4piepsi_(0))*(Qxxq)/((a//2)^(2))`
`implies q=(Q)/(4)`. hence `q=-(Q)/(4)`.