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Homework Help: Charges and Coulomb's Law

  1. Jan 30, 2010 #1
    1. The problem statement, all variables and given/known data

    Two positive point charges q are placed on the y-axis at a and -a. A negative point charge -Q is located at some point x on the +x-axis.

    Find the x-component of the net force that the two positive charges exert on -Q.

    Express your answer in terms of the variables q, Q, x, and a, with any necessary constants.

    2. Relevant equations

    Coulomb's Law:

    F = kq_1q_2/r^2

    3. The attempt at a solution

    The upper particle on the y-axis is q_1, the lower particle on the y-axis is q_2, and the particle on the x-axis is q_3.

    q_1 and q_2 have the same "pull" on q_3, so the vertical y-components of the force cancels out. Also, the force due to q_1 on q_3 is equal to the pull that q_2 has on q_3. The force on q_3 is equal to:

    F_q_3 = F_13 + F_23

    F_13 = kq_1q_3/r^2

    k = 8.99 * 10^9
    q_1 = q
    q_3 = -Q
    r^2 = a^2 + x^2 (pythagorean thereom to find the distance from q_1 to q_3)

    The magnitude of the force F_13 (particle one on particle three) is:

    F_13 = (8.99 * 10^9)(q)(-Q)/(a^2 + x^2)

    The x-component of this force is

    F_13x = (8.99 * 10^9)(q)(-Q)/(a^2 + x^2) * x/(sqrt(x^2 + a^2))

    Since F_q_3 = F_13 + F_23, and F_13 = F_23,

    F_q_3x = 2[(8.99 * 10^9)(q)(-Q)/(a^2 + x^2) * x/(sqrt(x^2 + a^2))]

    This is also what my teacher got for this problem, but it's incorrect. Can somebody please help me point out what the problem is?

    Thanks in advance.

    ADD: I've also tried:

    F_q_3x = 2[(8.99 * 10^9)(q)(Q)/(a^2 + x^2) * x/(sqrt(x^2 + a^2))]
    F_q_3x = 2[(q)(Q)/(a^2 + x^2) * x/(sqrt(x^2 + a^2))]
    F_q_3x = 2[(q)(Q)x/(a^2 + x^2)^(3/2)]
  2. jcsd
  3. Jan 31, 2010 #2
    I don't understand.

    Since I'm looking for the x-component of the force, I'd need the force directed along the x-axis, right? If the "height" of the triangle of one force is a, and the "base" is x, then the hypotenuse of the triangle would be sqrt(a^2 + x^2), right?

    Why would I need the vertical a value to solve for the x-component, especially when the vertical components cancel out?
  4. Jan 31, 2010 #3


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    Homework Helper

    Yes. You are right. Your answer is correct. Probably you have to include the sign, because the force in pointing towards the -ve x axis.
  5. Jan 31, 2010 #4
    Argh... I figured out what I did "wrong" in the end.

    I put that k = 8.99 * 10^9, so instead of writing down k, I wrote down the numerical value. And the system kept telling me that it was wrong. So the answer was:

    F_q_3x = 2[(k)(q)(-Q)/(a^2 + x^2) * x/(sqrt(x^2 + a^2))]

    That's incredibly maddening.

    Anyway, thanks for all the help. Now I'll need to ask my teacher to excuse the "mistake" that I made...
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