The charge of a ball?

NetherCraft 0

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!!!

1 Answer

  • 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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