Two identical point charges ,q each , are kept 2 m apart in air. A third pint charges Q of unknown magnitude and sign is placed on the line joining the charges such that the system remains in equilibrium. Find the position and nature of Q.

Explanation

Let two charges ,q each , are kept at points A and B respectively separated by a distance 2m. Let a third point charges Q be placed at point C situated at a distance x m from A .
For equilibrium of system the net force on any charges must be zero. Considering net force on charge Q, we have
` " "oversetto (F_(CA)) +oversetto (F_(CB)) =oversetto 0or |oversetto (F_(CA))|=oversetto (|F_(CB)|) `
` therefore " "(1)/(4 pi in _0) (qQ)/(x^(2))=(1)/(4pi in _0) .(qQ)/((2-x)^(2)) `
`rArr " " x= 2 -x or x=1 m `
Again considering net force on charge q situated at point A, we have
` " " oversetto (F_(AC)) +oversetto (F_(AB)) =oversetto 0 `
`rArr " "(1)/(4 pi in _0) .(qQ)/((1)^(2)) +(1)/(4 piin _0) .(q^(2))/((2)^(2)) =0 `
`rArr " " Q +(q)/(4) = 0 or Q= -(q)/(4)`