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Another charge question

  1. Jun 5, 2008 #1

    I have been reviewing a lot of electrostatic problems and I keep coming across problems that ask you to find the point where one would have to place a charge (between two other charges) such that the net force on said charge would be zero.

    I am trying work out a problem that I have with these, but I am not sure how to word what it is. Let me try though:

    Let us take two charges q1 and q2 of opposite sign. For simplicity's sake, let them be of equal magnitude. Let the 3rd charge, q3, be of equal magnitude as well. Clearly q3 must be placed such that it is collinear with the other two. Since F is a function of the distance r between them, there is only one spot in which the Net force on the q3 is zero.

    Now let us make the magnitude of the charge on q1 be twice that of q2. Let me make a simple sketch to make this easier:


    Let q3 be NEGATIVE and of magnitude q. It seems that it has to be placed to the right of q2 since as you approach from the right, from some large distance that makes d negligible, q1 is drawing q3 in towards it, but at some point the repulsion from q2 will counter that repulsion.

    Approaching from the left, q1 attracts q3 until it sticks to it, since the repulsive force between q2 and q3 will never be greater then the attractive force of q1 and q3.

    Now let q3 be POSITIVE and of magnitude q. If you place q3 at some large distance to the right it is drawn in by the attractive force of q2.... but this is where I get confused: How do I know if the repulsive force from q1 is large enough to counter the attractive force of q2?

    It seems like it could counter it but it would depend too many things, like d and the ratios of the magnitudes of the charges.

    Any input would be great here. Is this second case unanswerable without some more details ?
  2. jcsd
  3. Jun 5, 2008 #2


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    Hi Saladsamurai! :smile:

    I don't understand your confusion … negative to the left and positive to the right give essentially the same type of result.

    Since you understand the first, what worries you about the second? :smile:
    That's only three things … just write out the two forces, and find the value of d when the add to zero! :smile:
  4. Jun 5, 2008 #3
    I do not see how they are the same. The if the magnitude of the q1 and q2 are different and they are of opposite signs. So by changing the sign and the direction of approach of q3, there will certainly be different results.

    Re-read part 2. I can not see any other way for me to word it.... that is, how to word what my confusion is:redface:

    Thanks TT!~:smile:

    Here's the part

    Last edited: Jun 5, 2008
  5. Jun 5, 2008 #4


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    Hi Casey! :smile:

    If q3 is on the right, it is attracted to q2 with a force proportional to q2/r², and repelled by q1 with a force proportional to q1/(r+d)².

    They will be equal if 1 + d/r = √q1/√q2, or r = d√q2/(√q1 - √q2).

    So if q1 < q2, there can be no solution (for positive r).

    And if q1 > q2, there is always a solution. :smile:
  6. Jun 5, 2008 #5
    This makes sense! Thanks TT! This solves my confusion since there is an instance in which there is no solution. Perfect:smile:

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