electrostatics question(quick ans qiuck rate)?

Question

Two equal positive charges are separated by a distance 2a. For what value of y is the force on a charge Q is placed at point p maximum?
ans- y sq.rt2/2

Answer 1

I assume p is on the line bisecting the distance between the charges. The force from each charge is proportional to q*Q/r^2, where r is the distance from q to Q. That distance is the hypotenuse of a right triangle with legs a and y, so it is √[y^2 + a^2]. The force from each charge q is then K*q*Q/(y^2 + a^2). The x-component of forces from the charges are equal and opposite, so the cancel out, and the net force is in the y-direction only. The y-component of each force is F*y/√[y^2 + a^2] (the angle the force makes with the vertical is arctan (a/y) and the y-component is F*cos(arctan(a/y)). The total force is then

2*F*y/√[y^2 + a^2] = 2*K*q*Q*y*(y^2 + a^2)^-(3/2)

Take the derivative of F with resp to y to get

dF/dy = (2*K*q*Q)*[(y^2 + a^2)^-(3/2) - y*(3/2)*(y^2 + a^2)^-(5/2)*2*y


dF/dy = (2*K*q*Q)*(y^2 + a^2)^-(3/2) * [ 1 - 3*y^2*(y^2 + a^2)^-1]

set this to 0, leaving

1 - 3*y^2*(y^2 + a^2)^-1] = 0

y^2 + a^2 - 3*y^2 = 0

2*y^2 = a^2

y^2 = a^2/2

y = a/√2 = 2*a/√2