# The charge of a ball?

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A 5.0 nC point charge sits at x = 0. At the same time, a 4100 N/C uniform electric field (created by distant source charges) points in the positive x-direction.

At what point along the x-axis, if any, would a proton experience no net force?

At what point along the x-axis, if any, would an electron experience no net force?

Thanks for the help!!!

• Electrostatic force: F = kQq/r²

Electric field force: F = qE

Call the 5 nC charge Q

A proton must naturally lie on the negaive side of the x-axis. Since the

field will pull it in the positive direction while the force from Q pushes it

in the negative direction. To find the distance set these forces equal:

Fnet = qE – kqQ/r² = 0

qE = kqQ/r²

E = kQ/r²

r = √(kQ/E) = √[(9×10⁹Nm²/C²•5×10 ̄ ⁹C)/(4100N/C)] = 10.5 cm

=> Point along x-axis = -10.5 cm

The electron, since it has the same magnitude charge as the proton, will

experience the same forces from Q and the field. So it must also lie on the

same point on the negative x-axis. But the forces are reversed and the field

will instead push it to the negative side and Q will pull it towards the positive

x-axis. The net force for the electron would then look like:

Fnet = kqQ/r² – qE = 0

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