Two charges, ` + q and -q ` are separated by a distance d. At which points will the resultant electric field intensity be directed parallel to the line joining the charges ?

Explanation

Resultant electric field intensity at all points on the plane perpendicular to the line joining the two charges and passing through its mid - point , is directed parallel to the line joining the charges.
P is a point on this plane corresponding to the two charges at A and B. AP = BP = r ( say ).

Intensity at P due to the charge ` + q`.
` E _ 1 = ( q ) /( r ^ 2 ) ` , along ` vec ( PC) `
Intensity at P due to the charge ` -q`,
` E _ 2 = (q ) /( r ^ 2 ) ` , along ` vec ( PB) `
Therefore, ` E _1 = E _ 2 `
The resultant of these two intensities will be along ` vec (PD) ` , which is the bisector of the angle ` angle CPB`.
` therefore angle CPD = angle DPB = angle PBA = theta ` (say)
` therefore ` PD is parallel to AB.