Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Electric force problem (determining unknown position to make zero net force)

  1. Mar 9, 2009 #1
    1. The problem statement, all variables and given/known data
    A +2.0-nC charge is at the origin and a -4.0-nC charge is at x = 1.0-cm.
    At what x-coordinate could you place a proton so that is would experience no net force?

    [NOTE by me] I put q1 as the positive charge (+2.0nC) and q2 as the negative charge (-4.0nC). q0, of course, is the proton.

    2. Relevant equations
    - quadratic formula
    - F0 = F01 - F02 which becomes F01 = F02 when I replace F0 with zero (no net force).

    3. The attempt at a solution
    So I know that the proton must be either to the left of the positive charge or the right of the negative charge. It can't be in between these two charges for it to have zero net force.

    I plugged in numbers into this equation (F01 = F02) which became:
    (2 x 10-9)/(r2) = (4 x 10-9) / ((1.0 x 10-2) + r)2
    I simplified this and got:
    0 = r2 + (2 x 10-2)r - (1 x 10-4)

    Then I solved for r using quadratic formula and got:
    r = (+ 4.1421 x 10-3) or (- 2.4142 x 10-2).

    I thought the negative number would automatically cancel out because r is distance but the actual answer goes with the negative number. I'm confused at how cramster.com explained it.

    The book is called Physics for Scientists and Engineers A Strategic Approach (2nd) by Randall Dewey Knight) and its chapter 26 number 47. Here's a direct link (have to be a member though; it's free to look at odd numbers): http://physics.cramster.com/physics-for-scientists-and-engineers-a-strategic-approach-2nd-problem-9-718769.aspx [Broken]

    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Mar 9, 2009 #2


    User Avatar
    Homework Helper

    Well r is measured relative to the origin, so if you get r is positive, you place the proton where the force between the proton and the +ve charge pushes it towards the negative charge.

    Otherwise, r is negative, and the proton is placed to the left of the +ve charge. The force between the proton and the positve charge pushes it away i.e. towards the left. The Force between the proton and the -ve charge will attract it i.e. attract it towards the right. Thus with r negative, there exists a location where the net force is zero.

    Do you understand a little more now?
  4. Mar 10, 2009 #3
    Thanks for answering my question. I really appreciate it. And I'm sorry I still don't understand :(

    I understand that r can be negative and relation to the origin but I don't understand how to determine which number (that I found for r) is used as the answer.

    If r is positive, the proton will be on left side of the positive charge, right? I thought it would be attracted to the negative charge and repelled by positive charge to possibly get a net zero force, no?

    And same thing for negative r. It will be on the right side of the negative charge, right? Then it would be attracted to the negative charge and repelled by positive charge in the same way to possibly get a net zero force.

    Mm I don't know if I'm making my question clear. Do you have to determine the position by calculating it and prove if the net force is zero or not?

    Thanks again.
  5. Mar 10, 2009 #4


    User Avatar
    Homework Helper

    Since you used the condition [itex]F_{01} = F_{02}[/itex], whatever value of r you get, that is where the net force is zero. But I can't see how the book explained it as I need to join. I understand what you're saying and both answers should be correct.
  6. Mar 10, 2009 #5
    This is what it said:

    "x = +0.414cm or -2.41cm at the above two points the magnitudes of the two forces are equal. At point .414cm the magnitudes of both the forces are equal but the directions are same. But we require the forces to be in opposite direction. So the required position of the proton is 2.41cms on the side opposite q_2. "

    They also used same equation as me: F01 = F02.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook