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Electric Fields

  1. Apr 3, 2006 #1
    never mind. sorry :)
    Last edited: Apr 4, 2006
  2. jcsd
  3. Apr 4, 2006 #2
    I don't know where my calculator is, so this might be wrong... but the theory should be there.

    [tex] Qa [/tex]
    [tex] Qb [/tex]
    are point charges.

    [tex] \vec E = \frac{kq}{r^2}\hat r [/tex]

    [tex] \vec r_{PQa} = \left[\begin{array}{c} 11 \cross 10^2 m \\ 0 \\ 0 \end{array} \right][/tex]

    [tex] \vec r_{PQb} = \left[ \begin{array}{c} (11+22) \times 10^2 m \\ 0 \\ 0 \end{array} \right][/tex]

    [tex] r_{PQa} = |vec r_{PQa}| = \sqrt{(11\times 10^2 m)^2+(0)^2+(0)^2}=11\times 10^2m[/tex]

    [tex] r_{PQb}=33\times 10^2 m [/tex]

    [tex] \vec E_{net} = \vec E_{Qa} + \vec E_{Qb} = \frac{kQa}{r_{pQa}^2}\hat r_{Qa} + \frac{kQb}{r_{{pQb}^2}\hat r_{Qb} [/tex]

    Notice that:
    [tex] \hat r_{Qa}=\hat r_{Qb} = \hat i = \left[ \begin{array}{c} 1 \\ 0 \\ 0 \end{array} \right] [/tex]

    [tex] \vec E_{net} = k\hat i \left(\frac{4\times 10^{-6}}{(11 \times 10^2 m)^2} + \frac{-1 \times 10^{-6}}{33 \times 10^2 m}\right) [/tex]

    After you plug in [itex] k [/itex] and cross of the units you get:
    [tex] \vec E = \hat i \left((9\times 10^9) \frac{4\times 10^{-6}}{(11 \times 10^2 m)^2} + \frac{-1 \times 10^{-6}}{33 \times 10^2 m}\right)\fraq{N}{C}[/tex]

    Also remember that you can drop the vector notation for [itex] \vec E_{net} [/tex] by taking the magnitude of it. And since you only have one component [itex] E_{net} is just equal to whatever that calculation is above. Sorry, I really don't feel like doing that by hand... not that it's hard. I'm just too lazy to do it.

    For the second part of your question recall that:
    [tex] \vec F_{12} = \frac{kq_1 q_2}{r^2}\hat r [/tex]

    I think you can figure it out from what I showed you above. I gave a very rigrious calculation, not skipping many steps. I find it easier just to think about most questions like this as three dimensional questions, and just do a little bit more steps with the vector calculus.
  4. Apr 4, 2006 #3
    well.. the latex thing isn't working atm, i guess. ^^
    i solved the problem, however. i just didn't think about the types of the resulting charges.
  5. Apr 4, 2006 #4
    Why are you taking the absolute value? How are you setting up the problem for the force?
  6. Apr 4, 2006 #5
    Cool. Well I guess you're good.

    Maybe some day Latex will come back :)
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