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Two point charges, Zero Net Force

  1. Jun 7, 2008 #1
    [SOLVED] Two point charges, Zero Net Force

    1. The problem statement, all variables and given/known data
    Two point charges of charge 7.40 μC and -1.80 μC are placed along the x axis at x = 0.000 m and x = 0.300 m respectively. Where must a third charge, q, be placed along the x axis so that it does not experience any net force because of the other two charges?



    2. Relevant equations

    F = k[tex]\frac{Q1 Q2}{x^{2}}[/tex]


    3. The attempt at a solution

    im fairly sure i know what to do, but my answer is wrong. It basically looks like this:


    -k[tex]\frac{q1}{(0.3 - x)^{2}}[/tex] = k[tex]\frac{q2}{x^{2}}[/tex]


    and then i just fill in the two charges, (coulomb's constant cancels) and solve for X. It comes out to a quadratic equation which isnt a big deal; my positive answer is 2.41E-1 m (which is incorrect).



    heres a diagram:
    [​IMG]
     
  2. jcsd
  3. Jun 7, 2008 #2

    alphysicist

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    Hi k3nt70,

    The first think to do in these problems is decide where the third charge goes: between the charges, on the left, or on the right.

    Let's assume the third charge is positive, and place it in the middle of the other two. What direction will the force from the 7.4 charge point, and what direction will the force from the -1.8 charge point? They'll both be to the right, so there's no way that they can cancel (we need the forces to be in opposite directions). (If the third charge were negative, both forces would be to the left, so the conclusion would be the same.)

    So it will either be on the left or the right. How do you think you can choose which side it's on?

    Once you get that, your basic approach of setting the two force magnitudes equal to each other is correct; you'll just have to adjust the denominators for the new situation. (And you don't want a minus sign, since we are setting the magnitudes equal.)
     
  4. Jun 7, 2008 #3
    i see. So it would have to be a positive or negative charge to the right of the -1.8 charge. Does that mean it should look like this:

    [tex]\frac{Q1}{(0.3 + X)^{2}}[/tex] = [tex]\frac{Q2}{X^{2}}[/tex]

    if so, then it has only changed my answer to a negative which is still incorrect.
     
  5. Jun 8, 2008 #4

    alphysicist

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    What numbers were you putting into your equation? Remember that in coulomb's law for force and fields we use the magnitudes of the charges.

    What do you get for x? And remember, once you find x, that's not quite the final answer they are asking for.
     
  6. Jun 8, 2008 #5
    so, it breaks down into...:

    5.6E-6(X^2) + 4.44E-6(X) + 6.66E-7

    This makes my final answers come out to two negative numbers, neither of which can be correct, can they?

    -0.20091
    -0.59195
     
    Last edited: Jun 8, 2008
  7. Jun 8, 2008 #6
    9.2E-6(X^2) + 4.44E-6(X) + 6.66E-7

    These are the numbers i was originally using; this is with a positive 1.8 value. It still yeilds the wrong x value :(
     
  8. Jun 8, 2008 #7

    alphysicist

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    Can you show the steps on how you got this? The coefficient of the first term (the x^2 term) looks right, but I get different answers for the other two.
     
  9. Jun 8, 2008 #8
    sure.

    1.8E-6(X^2) = 7.4E-6((0.3+X)^2)
    1.8E-6(x^2) = 7.4E-6(0.09 + 0.6X + X^2)
     
  10. Jun 8, 2008 #9

    alphysicist

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    Oh, that explains it. You have Q1 and Q2 reversed; from your equation Q2 is the charge that is a distance X away, and so Q2 is the 1.8 charge and Q1 is the 7.4 charge.
     
  11. Jun 8, 2008 #10
    ohh. ok, so my quadratic is:
    -5.6X^2 + 1.62X +1.8.
    This gives:
    -0.44046
    0.72975

    Now i recall you saying there was an additional step after ive found the x-value. I have no idea what that might be lol.
     
  12. Jun 8, 2008 #11

    alphysicist

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    I think you still have a math error somewhere. Let me take your last equation and switch the charges:

    7.4E-6(x^2) = 1.8E-6(0.09 + 0.6X + X^2)

    Expanding:

    7.4E-6(x^2) = 0.162 + 1.08 x + 1.8E-6 x^2

    So the last two terms in your quadratic need to be 0.162 and 1.08. What does the quadratic equation then give you?

    The additional step I was talking about was that they want the displacement from the 7.4 charge (at the origin). As you said, we are solving for x which is the distance from the -1.8 charge, so once you find x you'll have to account for that.
     
  13. Jun 8, 2008 #12
    -0.099089
    0.29195.

    So... 0.29195 + 0.3 = 0.59195 m

    Does that seem about right?
     
  14. Jun 8, 2008 #13

    alphysicist

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    That sounds right to me.
     
  15. Jun 8, 2008 #14
    woohoo! 100% on this assignment. THanks alot for your help!!
     
  16. Jun 8, 2008 #15

    alphysicist

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    Sure, glad to help!
     
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