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Electric field

  1. Sep 3, 2005 #1
    Can someone help with this? I want to say the answer is zero but I don't know how to explain it.

    Calculate the electric field at the origin due to the following distribution of charges: +q at (x,y)=(a,a), +q at (-a,a), -q at (-a,-a) and -q at (a,-a).
     
  2. jcsd
  3. Sep 3, 2005 #2
    Each charge has an E field described by

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

    Just find the vector sum at the origin, or better yet draw it out and see if you can cancel by symmetry.
     
  4. Sep 3, 2005 #3
    thanks

    Thank you for the reply.
    would i be correct is saying that

    E= (+q*k/a^2+a^2)+(+q*k/a^2+a^2)+(-q*k/a^2+a^2)+(-q*k/a^2+a^2)= 0

    ??
     
  5. Sep 4, 2005 #4
    Your conclusion is correct, but you measured the distance from the origin incorrectly, remember the distance between the origin and the point (x,y) is

    [tex] \sqrt{x^2+y^2} [/tex]
     
  6. Sep 4, 2005 #5
    thank you

    thanks again.
    in this case wouldn't the distance from the origin be (sq rt of (a^2+a^2))? In the original problem all the charges are at (a,a),(-a,a),(-a,-a) and (a,-a).
     
  7. Sep 4, 2005 #6
    Yes thats correct, it simplifies to [itex] \sqrt{2}a[/tex]
     
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