Assume that a ball of charged particles has a uniformly distributed negative cha

Question

Assume that a ball of charged particles has a uniformly distributed negative charge density except for a narrow radial tunnel through its center, from the surface on one side to the surface on the opposite side. Also assume that we can position a proton anywhere along the tunnel or outside the ball. Let FR be the magnitude of the electrostatic force on the proton when it is located at the ball's surface, at radius R. As a multiple of R, how far from the surface is there a point where the force magnitude is 0.71FR if we move the proton (a) away from the ball and (b) into the tunnel? (a) Number Units Units (b) Number

Explanation / Answer

A)

Electric field, E = (1/ 4pi e0) (q/r^2)

Force, F = qE

0.71 Fr = (1/ 4 pi e0) (q qp/r^2)

0.71/R^2 = 1/r^2

r^2 = (1/0.71)R^2

r = 1.187 R

So, distance, d = 1.187 R - R = 0.187 R

B)

E = (1/4 pi e0) (q/R^3)r

F = qE

0.71Fr = (1/ 4 pi e0) (q qp/R^3)r

0.71 = r/R

r = 0.71 R

Distance, d = R - 0.71 R = 0.29 R

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