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Coulomb's Law problem

  1. Mar 16, 2009 #1
    1. The problem statement, all variables and given/known data

    Particle 1 of charge q1 = +0.76 µC and particle 2 of charge q2 = -3.0 µC, are held at separation L = 13 cm on an x axis. If particle 3 of unknown charge q3 is to be located such that the net electrostatic force on it from particles 1 and 2 is zero, what must be the x and coordinates of particle 3?

    I'm really not sure how to proceed with this question. But I understand that the forces added together is zero.
     
  2. jcsd
  3. Mar 16, 2009 #2

    LowlyPion

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    Welcome to PF.

    OK. So what equation do you think you might use then?
     
  4. Mar 16, 2009 #3
    So I did:

    c/d^2 + c/d^2 =0

    put in the numbers

    (0.76*10^-6)/d^2 = (3*10^-6)/(d+13)^2

    simplify taking out the fractions

    (0.76*10^-6)(d+13)^2 = d^2(3*10^-6)

    expand the bracket

    (0.76*10^-6)(d^2+26d+169) = d^2(3*10^-6)

    eventual rearrangement

    2.24*10^-6 - 1.976*10^-5 - 1.2844*10^-4 = 0

    quadratic equation

    13 and -4.3 but 13 is not the right answer. It might be my method or calculation error, I'm really not sure.
     
  5. Mar 16, 2009 #4

    LowlyPion

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    I trust you didn't round away too much from the answer for your quadratic.

    If you think about it, ≈ 13 is an expected result, because if 1 charge is 4 times larger and twice as far away ... you'd expect something about equal to the charge separation wouldn't you?
     
  6. Mar 16, 2009 #5
    Hum, isn't q2 negative? I reckon you might've used it as a positive..

    F1 - F2 = 0 .. so, q1/d^2 = q2/(d+x)^2, where q2 still is negative, and I only see +3 in your calculations :)
     
  7. Mar 16, 2009 #6

    LowlyPion

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    Actually I think he's taken that into account.

    ∑F = 0 = q3*∑ E = 0 ⇒ E1 + E2 = 0

    Taking account of the sign of the charge then |E1| = |E2| satisfies the condition.

    (This of course is for x that does not lie between q1 and q2.)
     
  8. Mar 16, 2009 #7
    I don't quite see it, but I reckon you're right 'cause you most definitely know better than I do and it's late & I'm tired. I think I'll come back and try to see it when it's not 20 to midnight ;)
     
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