1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    rl.bhat

    User Avatar
    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...
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Charges and Coulomb's Law
  1. Coulomb's law charge (Replies: 1)

Loading...