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Coulomb's law and vectors

  1. Sep 6, 2004 #1
    To find the magnitude of a force between two charges is very simple, but to get the direction of the force seems a little strange to me. The signs of the charges aren't included anywhere in the law, so does this mean that I literally have to think "the signs are opposite, therefore I will multiply the vector by -1 (or leave it positive, depending on my reference coordinates)"? This seems a little bit cumbersome and forced to me, and apparently not derived anywhere in the mathematics of it. Am I going about doing this the right way?

    I'm quite accustomed with vectors (vector calculus and linear algebra background), but very new to e&m. Any help would be much appreciated.
  2. jcsd
  3. Sep 6, 2004 #2


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    The signs are included in Coulomb's law.
  4. Sep 6, 2004 #3
    "The signs of the charges are not included anywhere in the law". Hmmmmmmmmmmm.
    I have a good book with an equation that should help you out.

    F= C q1 q2 / r12^2 * (r12)/r12

    Where the bolds are vectors, the regular font is scalar
    and r12 =r1-r2
    where r1 is the location of the q1 and
    r2 is the location of q2

    If I'm not too clear, I'll fix it up some more
  5. Sep 6, 2004 #4


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    The "force on q1 due to q2" is (in agreement with sinyud)
    [tex]\vec F_{on\ q_1\ due\ to\ q_2} = k\frac{q_1 q_2}{r_{12}{}^2} \hat r_{12}[/tex]
    where [itex]\hat r_{12}[/itex] is the unit vector at the target charge [itex]q_1[/itex] pointing away from the source charge [itex]q_2[/itex] and
    [itex]r_{12}[/itex] is the distance to the target charge [itex]q_1[/itex] from the source charge [itex]q_2[/itex].

    If the product [itex]q_1q_2[/itex] is positive (so they have like signs), then,
    since [itex]\hat r_{12}[/itex] points away from [itex]q_2[/itex], it follows that[itex]\vec F_{on\ q_1\ due\ to\ q_2}[/itex] points away from [itex]q_2[/itex].
    "[itex]q_1[/itex] is repelled by [itex]q_2[/itex]."

    If the product [itex]q_1q_2[/itex] is negative (so they have unlike signs), then,
    since [itex]\hat r_{12}[/itex] points away from [itex]q_2[/itex], it follows that [itex]\vec F_{on\ q_1\ due\ to\ q_2}[/itex] points towards [itex]q_2[/itex].
    "[itex]q_1[/itex] is attracted to [itex]q_2[/itex]."
  6. Sep 6, 2004 #5
    Ahh, thanks guys. For some reason I was attaching the unit vector to the equation for the magnitude of the force, which obviously removes the signs from the charges. Major brain fart.

    Thanks again.
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