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Coulomb's Law

  1. Jan 22, 2004 #1
    Hello all,

    I have two charges, q1 & q2, along a horizontal axis of length L. I supposed to determine the length at which another charge, Q, can be placed so that its net force is zero, other than infinitely away.

    I'm using the equation:
    F = kq1Q / r + kq2Q / r
    -- > kq1Q / (L+x)^2 = -( kq2Q / x^2 ) where x is the distance the point is away from L.

    I substitute my #s and I get the equation:
    (8 / (l^2 + 2xL + x^2)) = -(2/x^2)
    and then I get:
    8x^2 = -2(L^2 + 2xL + x^2)
    -4x^2 = L^2 + 2xL + x^2
    0 = L^2 + 2xL + 5x^2
    Is this right? I cannot seem to solve for x.

    I thought I initially had it, but reworked through my math and realized I forgot to have 2q negative. What I initally got was:
    L^2 + 2xL - 3x^2
    I can solve that by 'unfoiling' easily, but, as I previously mentioned, the math isn't correct to get that equation.

    Thank you for your help and have a wonderful day.
     
  2. jcsd
  3. Jan 22, 2004 #2

    Doc Al

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    Staff: Mentor

    What values are you using for q1 and q2?
     
  4. Jan 22, 2004 #3
    I'm using 8 for q1 and -2 for q2.
     
  5. Jan 22, 2004 #4

    Doc Al

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    Staff: Mentor

    What makes you think the math isn't correct? Looks right to me.
     
  6. Jan 22, 2004 #5
    I start with this equation:
    (8 / (L^2 + 2xL + x^2)) = -(-2/x^2), cross multiply
    8x^2 = -2(L^2 + 2xL + x^2), divide by -2
    -4x^2 = L^2 + 2xL + x^2, add -4x^2
    0 = L^2 + 2xL + 5x^2, not L^2 + 2xL - 3x^2 as I had originally hoped for.
    So, how do you solve for x (in terms of L) for this equation:
    0 = L^2 + 2xL + 5x^2
     
  7. Jan 22, 2004 #6

    Doc Al

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    Staff: Mentor

    You are messing up with signs. In your first equation, the minus signs cancel. Your original hope was correct. :smile:
     
  8. Jan 22, 2004 #7
    I was in the middle of asking a question about the signs and your response answered it perfectly.

    Thank you.
     
    Last edited: Jan 22, 2004
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