Question

Two positive charges of 1.5e-10 C are placed on the x axis at distances +12 cm and -12 cm from the origin, respectively. A third positive charge of 5e-11 C can be moved along the y axis. At what value of y the electrostatic force on the third charge is maximal?

Answer 1

Let's call the two stationary charges q, and the other charge Q. Call the distance from q to the origin L.

Now for any point on the y-axis y, the force on Q has two components - one in the x direction and on in the y. But by the symmetry of teh problem, the force in the x direction is always zero. So we only need to look at the force in y. The magnitude of teh force due to each charge q is:

F = kqQ/r^2 where r = sqrt(y^2 +L^2)

Now the component of force in the y-direction is given by the cosine angle, a, each force vector makes witht he y-axis and the magnitude of teh force, F

Fy = kqQ/r^2 cos(a) But cos(a) can be found from the geometry of the problem

cos(a) = y/sqrt(y^2+L^2) so

Fy = kqQ y/(y^2+L^2)^(3/2)

Since there are two charges q, the total force in y is

Fty = 2*Fy = 2*kqQ y/(y^2+L^2)^(3/2)

Now the maximum force occurs when

dFty/dy = 0 = 2kqQ/(y^2+L^2)^(3/2) - 6kqQ y^2/(y^2+L^2)^(5/2)


0 = 1 - 6y^2/(y^2+L^2) ---> 6y^2 - y^2 - L^2 = 5y^2 - L^2 = 0 ---> y = +/- L/sqrt(5)

Here we have given L = 12 cm

y =  +/- 12/sqrt(5) = +/- 5.36 cm -------------------------------Ans.