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Homework Help: Point Charges and Force vector addition

  1. Sep 15, 2008 #1
    Hi. I would appreciate some help with this point charge problem. I think I understand the general set up, but don't know where to go from there.

    1. The problem statement, all variables and given/known data

    Two point charges are placed on the x axis as follows: Charge q1 = +4.00nC is at x=0.400m, and charge q2 = +2.00nC is at x=-0.200m. State, in x and y coordinates, where a negative point charge, q3 = -4.73nC, must be placed to have a zero net force exerted on it by q1 and q2.

    2. Relevant equations

    I know that F = (k*q*Q)/r2 and that the force from q1 acting on q3 (F13) has to be equal to the force from q2 acting on q3 (F23) for the net force to = 0.

    3. The attempt at a solution

    Where I am getting lost is in relation to the distances between the charges. If I don't know the distance between q1 and q3, and q2 and q3, how can I do the problem? And I can't rely on the Pythagorean theorem because there is no guarantee that a right angle is formed. I know my best chance at solving this is a quadratic equation, but I don't know how to get to that point. It has been a LONG time since I took algebra, so I may be overlooking something very simple. This is what I have:

    F13 = (k*q1*q3)/r132 = F23 = (k*q2*q3)/r232

    things cancel and it simplifies to

    q1/r132 = q2/r232

    From here I have absolutely no idea where to go. Thanks for your help.
  2. jcsd
  3. Sep 16, 2008 #2


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    The approach you should take in this question is to ignore the magnitude and sign of the negative charge to be placed, and determine, from the static charge placement of the other two charges which spot on the plane has the resultant E-field is zero. By symmetry, this spot should lie on the x-axis. Now just use Coulomb's law in one dimension to find the distance to required spot.

    EDIT: I'm assuming q1 and q2 are fixed and unable to move.
  4. Sep 17, 2008 #3
    I think I got the problem, but can someone check my work? Thanks!

    Because q1 and q2 are both positive, the only place that you can put q3 to make fnet = 0 would be between them on the x axis.

    F23 = F13
    Kq1q3/(r1-x)^2 = kq2q3/x^2
    q1/(r1-x)^2 = q2/x^2
    q1x^2 = q2(r1-x)^2
    q1x^2 = q2(r1^2 - 2r1x + x^2)
    q1x^2 = q2x^2 - 2q2r1x - q2r1^2

    plugging in the numbers,

    4.00E-9x^2 - (2.00E-9)(0.6)^2 + (2.00E-9 * 0.6)x - (2.00E-9)x^2 = 0
    2.00E-9x^2 +2.4E-9x - 7.2E-10 = 0
    by the quadratic formula:
    x = -2.4E-9 +/- sqrt((2.4E-9)^2 - 4(2.00E-9)(-7.2E-10))/ 2(2.00E-9)
    x = 0.249, -1.45
    Because -1.45 is not between the q1 and q2 it can be discarded.
    So x is 0.239m away from q1, so 0.400- 0.249 = 0.151. So the coordinates are (0.151,0).
    Last edited: Sep 17, 2008
  5. Sep 18, 2008 #4


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    This line has a sign error in the last term (but it looks like you correct for that in the next line).

    The third term is missing a factor of 2, but the rest is okay. (It looks like you fix the missing factor of 2 in the next line.)

    These results look right.

    No, that is not correct. If you look at your starting equation:

    F23 = F13
    Kq1q3/(r1-x)^2 = kq2q3/x^2

    x is the distance from q2, not q1. So at what x coordinate would the third charge have to be placed at?
  6. Sep 18, 2008 #5
    Would this then be 0.6-0.249 = 0.351?

    I'm kind of confused because I keep doing the quadratic equation for this problem over and over, and one way I did it (plugging into calculator and solving) I got x = 0.249, then when I expanded it and did it by hand, I ended up getting 0.351.

    4/(x(13))^2 = 2/ (0.6 - x(13)^2)
    4(x(13)^2 - 2x(13) + 0.36) = 2x(13)^2
    The final equation I got was 2x(13)^2 - 4.8x + 1.44 = 0
    When I do the quadratic for this, I end up getting 0.351.

    I'm not really sure what part of this I'm doing wrong, & why there is a discrepancy between my answer the one that my calculator is showing.

    Thanks :)
  7. Sep 18, 2008 #6


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    This line is correct, but the first two lines have some errors (though they may just be typos?). This equation will give x=0.351, as you found. What equation did you put in your calculator to give you 0.249?

  8. Sep 18, 2008 #7
    I'm not really sure what I plugged in, because now that I'm doing it again, i really do get x = 0.351, haha.

    So now I'm just trying to figure out the coordinate, which I think would be to take this x that i solved for subtract from the total distance, and then subtract it from the distance of q1 and add it to the distance of q2 both from the origin).
  9. Sep 18, 2008 #8


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    Can you post your work? I don't think that procedure is correct, but maybe I'm misreading your post.

    I think the important thing is first to think about this: You found that the distance of the test charge is 0.351m; but what does that mean? The test charge is 0.351m from what?

    Then if they want the x-coordinate, that means to just find the distance the test charge is from the origin, with the coordinate being positive if it's on the right and negative if it's to the left of the origin. (assuming your using the normal convention for the signs)
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