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Electric Field question

  1. Jan 6, 2005 #1
    Electric Field question:

    Two charges are located on the x axis: q1 = +6.0 micro-Coloumbs at x1=+4cm, and q2=+6 micro-coloumbs at x2 = -4cm. Two other charges are located on the y-axis: q3 = +3 micro-Coloumbs at y3=+5cm, and q4 = -8 micro coloumbs at y4 = +7cm. Find the net electric field (magnitude and direction) at the origin.

    What I've tried to do is say E = F/q, and F = Kq1q2 / r^2, therefore E = (kq1q2 / r^2) / q. since q2 = origin, it = 1? and the q1 and q cancel out? So it becomes: E=k/r^2?

    I'm pretty sure i'm doing it wrong and was wondering if someone could just point me in the right direction?

    Any help appreciaited, thanks. :smile:
     
  2. jcsd
  3. Jan 6, 2005 #2

    dextercioby

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    HINT:The electric (electrostatic) field created by a charge 'q' at the point [itex] \vec{r} [/itex] is given by
    [tex] \vec{E}_{q,\vec{r}}=:\frac{q}{4\pi\epsilon_{0}r^{2}}\frac{\vec{r}}{r} [/tex]
    ,where [itex]\vec{r} [/itex] is the position vector at the point u wish to calculate the fiels wrt to the point in which is the source "q".

    Daniel.

    PS.You'll have to apply that formula 4 times (for each charge) and then add those 4 vectors obtained.
     
    Last edited: Jan 6, 2005
  4. Jan 6, 2005 #3
    E = F/q(o) where q(o) is the charge experiencing the force.

    so E = Kq1q(o)/(r^2*q(o)) <----q1 is the charge providing the force and the q(o) is the charge that would be at the origin. However though, as you mentioned before, it does cancel out.
     
  5. Jan 6, 2005 #4
    Thanks for the help everyone.
     
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