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Homework Help: How would you have done this Coulomb's law problem?

  1. Jul 22, 2012 #1
    1. The problem statement, all variables and given/known data

    In the figure, particle 1 of charge 1.0 micro Coulombs and particle 2 of charge -3.0 micro Coulombs are held at a separation of L = 10.0 cm. If particle 3, of unknown charge is to be located such that the net electrostatic force on it is 0, what must its x and y coordinates be?

    (figure)

    1<------L------->2

    with 1 at the origin.


    2. Relevant equations

    Coulomb's law

    3. The attempt at a solution

    First, I noted that the particle must be to the left of 1 because 2 is of stronger net charge, so it must be closer to 1 and further from 2.

    Furthermore, since the sum of the two forces is 0, they must be equal in magnitude and opposite in sign.

    [itex]kq_{1}q_{3}r^{-2}_{1}\widehat{r}_{1} = - kq_{2}q_{3}r^{-2}_{2}\widehat{r}_{2}[/itex]

    [itex]q_{1}r^{-2}_{1}\widehat{r}_{1} = - q_{2}r^{-2}_{2}\widehat{r}_{2}[/itex]

    I was stumped here until I cheated and looked at the y coordinate answer. It is 0. Only knowing that was I able to solve. Can someone tell me why that is the case?


    Since I know the y coordinate is 0, the vectors on both is -i, allowing me to cancel them out.


    [itex]q_{1}r^{-2}_{1} = - q_{2}r^{-2}_{2}[/itex]

    [itex](1x10^{-6})r^{-2}_{1} = (3x10^{-6})r^{-2}_{2}[/itex]

    [itex]r^{-2}_{1} = 3r^{-2}_{2}[/itex]

    [itex]r^{2}_{2} = 3r^{2}_{1}[/itex]

    [itex]r_{2} = \sqrt{3}r_{1}[/itex]

    I also know that r2 is 10 greater than r1.

    [itex]r_{2} = 10 + r_{1}[/itex]

    Substituting that for r2 in the first equation,

    r1 = 13.66.

    Since I know it's to the left of r1, and r1 is on the origin, the x coordinate must be -13.66.

    The answer in the book is -14, so I'm reasonably confident.

    However, I want to know:

    1.) Is this how you would have done it?
    2.) Why is the y component 0?
    3.) Is the y component known to be 0 through working the problem, or from examining the layout of the particles?

    TY
     
  2. jcsd
  3. Jul 22, 2012 #2

    Doc Al

    User Avatar

    Staff: Mentor

    Hint: Can two non-parallel vectors cancel?
     
  4. Jul 22, 2012 #3
    Nope, I see. Thanks.

    does the rest look ok?
     
  5. Jul 22, 2012 #4

    Doc Al

    User Avatar

    Staff: Mentor

    I didn't check your arithmetic, but your method is perfectly correct.
     
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