1. Not finding help here? Sign up for a free 30min 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!

Electric field and charges

  1. Feb 25, 2007 #1
    1. The problem statement, all variables and given/known data

    Two point charges are on the y-axis. One, with charge a of +q, is located at point a. The other, with a charge of +2q is located at point -a. Determine the magnitude and direction of the electric field at the origin. Express answers in terms of q,a, and constants.

    2. Relevant equations


    3. The attempt at a solution

    Do i just do k(2q)/a^2 - k(q)/a^2 and thats the answer?

    K(2q)/a^2 - k(q)/a^2= kq/a^2

    this in preparation of the ap exam and i learned this last semester so i just need a little refreshing
  2. jcsd
  3. Feb 25, 2007 #2
    Yes, and the direction is the direction a positive test charge would go
  4. Feb 25, 2007 #3


    User Avatar
    Gold Member

    Yup, thats how I would do it.

    Since both are a distance "a" from the origin and we only care about what happens at the origin, you equations makes sense to me.

    EDIT: oops, turdferguson beat me to it.
  5. Feb 25, 2007 #4
    the direction is in the -y direction right?

    also, if i were to find the electric potential at the origin i would use E=V/d? if i use that, im not sure what i would use for d

  6. Feb 25, 2007 #5
    Think about the direction again, all charges are positive

    d is just the distance a. A good thing to remember for the AP is that fields are vectors and potentials are scalars. Voltage is also the scalar sum of each kq/r
  7. Feb 25, 2007 #6
    ok so its in the positive y direction because the the electric field points outward from positive charges and since the 2q charge is greater the field at the origin will point in the positive y direction
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Electric field and charges