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Homework Help: Coulomb's Forces

  1. Apr 15, 2014 #1
    1. The problem statement, all variables and given/known data

    Consider the three charges in the figure below, in which d = 4.7 cm, q = 11 nC, and the positive x-axis points to the right. What is the force Fvec on the 5 nC charge? Give your answer as a magnitude and a direction.

    2. Relevant equations

    F= (K*Q*q)/r^2

    3. The attempt at a solution

    First I found the distance between +5 and 11.
    (.047^2)+(.03^2)=.055758 m

    arctan(.047/.03)= 57.45

    Then the magnitude: .000164

    The the magnitude of +5, -5 --> .00025 (Horizontal component)

    cos 57.45 = x/.000164 = 8.823X10^-5 (Vertical component of line)

    sin 57.45 = x/.000164= 1.3824x10^-4 (Horizontal component of line)

    Add horizontal components and use that and vertical to make a triangle and find the third side.

    So my answer is incorrect, can someone help me out please. Thank you.

    Attached Files:

  2. jcsd
  3. Apr 15, 2014 #2


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    Is the force Between the two positive charges attractive or is it repulsive ?
  4. Apr 15, 2014 #3
    Repulsive. I was unclear, but if the question is whether I took it into account when adding the horizontal components, I did. -->.00025 + (-1.3824x10^-4)
  5. Apr 16, 2014 #4
    Many of the equations you wrote are incorrect even if the solution is correct. That makes your solution very hard to understand. For instance, you wrote
    cos 57.45 = 8.823X10^-5 which is completely wrong. There are at least 3 incorrect things with that line and I see at least 3 other lines with similar mistakes.
  6. Apr 16, 2014 #5


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    dauto makes an excellent point.

    Look at your line:
    cos 57.45 = x/.000164 = 8.823X10^-5 (Vertical component of line)​
    Is x (in this case) supposed to be the vertical component of the force that the 11 nC charge exerts on the 5 nC charge at the origin ?

    If so you should use the sine, not the cosine. Even then, it's only the absolute value of that component.

    As dauto points out, what you have written says that cos(57.45°) = 8.823X10^-5 . I doubt that you mean that.

    Write something like:
    cos(57.45°) = x/.000164 → x = 8.823X10^-5​
    (Of course this is not the vertical component.)

    It wouldn't hurt to include units in your answer.
  7. Apr 16, 2014 #6


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    ok, you're given the distance, and you have the correct formula for your force, why not start with a N2L equation?
    [itex]\Sigma \vec{F}=m_{5}\vec{a_{5}}=K[\frac{q_{-5}q_{5}}{.03^2}\hat{i}+\frac{q_{11}q_{5}}{r_{11→5}^2}\hat{r}][/itex]
    [itex]\hat{r}=cos(tan^{-1}(\frac{d}{.03}))\hat{i} + sin(tan^{-1}(\frac{d}{.03}))\hat{j} [/itex]
  8. Apr 16, 2014 #7


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    I think I made a mistake, the i hat's going to give you a force in - i, that should be rhat_{-5->5} and the second should be rhat_{11->5}
    Last edited: Apr 16, 2014
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