1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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

    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]
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook