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)/(4)`
B. `(-Q)/(4)`
C. Q
D. `(-Q)/(8)`

Answer

B

Explanation

Let the distance between the charge q and Q be . System is in equilibrium then net force on q is zero
`therefore` Total force on Q at B is
`(KQq)/(x^(2))+(KQ Q)/((2x)^(2))=0`

`(KQq)/(x^(2))=(KQ^(2))/(4x^(2))`
`q=(-Q)/(4)`