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Homework Help: 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

    E=kq/r^2

    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

    ranger

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    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

    thanks
     
  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
     
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