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Homework Help: Calculating net force on a charge placed at a spot

  1. Jan 25, 2009 #1
    1. The problem statement, all variables and given/known data

    Two charges, -16 and +3.4 mC, are fixed in place and separated by 4.4 m. (a) At what spot along a line through the charges is the net electric field zero? Give the distance of the spot to the positive charge in meters (m). (Hint: The spot does not necessarily lie between the two charges.) (b) What would be the force on a charge of +14 mC placed at this spot?


    2. Relevant equations

    E=kq/r^2


    3. The attempt at a solution


    I've figured out part a, but I don't know what I'm doing wrong on part b... here is my work:

    a) the correct answer is 3.76 m.

    b) F21=(8.99x10^9)(3.4x10-6)(14x10-6)/(3.76 m)^2 = 0.03026 N
    F23=(8.99x10^9)(-16x10^-6)(14x10-6)/(4.4+3.76)^2 = -0.0302 N

    then I added the 2 together and got 1.68x10^-5 N
    Am I supposed to actually get 0 as my answer? I'm so confused.
     
  2. jcsd
  3. Jan 25, 2009 #2
    F=qE. Because E = 0 at this spot any charge placed at this spot will not experience a force. It's analogous to zero gravity; any mass placed in a zero gravity field will not experience a force.
     
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