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Homework Help: Electric Fields & Force

  1. Mar 13, 2008 #1
    A small object carrying a charge of -5.50 nC is acted upon by a downward force of 24.0 nN when placed at a certain point in an electric field.

    What would be the magnitude and direction of the force acting on a proton placed at this same point in the electric field?

    Magnitude & Direction of the Electric Field = 4.36 N/C upward

    F = k (Q1Q2/r^2)
    F = |q|E
    k = 9E9

    F = |q|E
    = 5.5E-9 x 4.36
    = 2.398E-8
  2. jcsd
  3. Mar 14, 2008 #2
    "Magnitude & Direction of the Electric Field = 4.36 N/C upward"

    I think this is correct so far, and then all you need to do is multiply the E by the charge of a proton.

    There is no need for k.

    I think u pretty much got it, maybe # error.
  4. Mar 14, 2008 #3

    Shooting Star

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    If you have found the field yourself, then what do you think should put in place of q?
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