# Answer to Question #21865 in Electromagnetism for divya rawal

Question #21865

two equal point charges A and B are R distance apart. A third point charge placed on the perpendicular bisector at a distance 'd' from the centre will experience maximum electrostatic force when d equals to what?

Expert's answer

Force of interaction with charge A q is F1 = kqQ/r^2 where r is the distance between q and Q. The total force is

F = 2*F1*cos\alpha = 2*F1*d/r = 2kqQd/r^3.

r = sqrt(d^2 + R^2/4).

Differentiate F by d: F' = 2kqQ* 8(R^2-8d^2)/(R^2 + 8d^2)^5/2 = 0 => d = R/2sqrt(2).

F = 2*F1*cos\alpha = 2*F1*d/r = 2kqQd/r^3.

r = sqrt(d^2 + R^2/4).

Differentiate F by d: F' = 2kqQ* 8(R^2-8d^2)/(R^2 + 8d^2)^5/2 = 0 => d = R/2sqrt(2).

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