- 11

- 0

Two small beads having positive charges 19q and q are fixed at the opposite ends of a horizontal insulating rod extending from the origin (the location of the larger charge) to the point x = d. As in the figure below, a third small charged bead is free to slide on the rod.

So I figure that

F = ((k * 19q Q)/ x^2) - (k* q * Q)/(d-x)^2)

so 19/x^2 = 1/(d-x)^2

d-x = x / sqrt(19)

d = x + x/sqrt(19)

= x * (1+sqrt(19))/sqrt(19)

so x = d * (sqrt(19))/(1 + sqrt(19))

which would be about 0.81339

apparently it's the wrong answer. Can anyone tell me why, or where I went wrong?