Question

two equal point charges are kept a distance R apart. A third charged particle kept on the perpendicular bisector at a distance r form the center will experience maximum electrostatic force

Answer 1

Answer:

Third charge on bisector of charge at distance R so distance of third charge from both charge is √(R^2/4+d^2),

direction of force along the side of triangle formed by these charge, for on due to these both chages at algle and angle can be easily calculate and their horizontal components cancel our only vertical component remains and this is maximum force on third charge,