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: Expression of Electric Field with known charges

  1. Jul 20, 2010 #1
    1. The problem statement, all variables and given/known data

    I have 2 charges:
    Q1 = 300 uC located at (1, -1, -3) m
    Q2 = ?? uC located at (3, -3, -2) m

    Q1 feels a force from Q2 of F1=(8i - 8j + 4k) N

    1. Find Q2
    2. Find the expression of the E-Field at (3, -3, -2) m (at Q2)

    2. Relevant equations
    F=(k|Q1*Q2|) / r^2
    E=(k|q|) / r^2

    3. The attempt at a solution
    For the first part I've used the magnitudes:
    Q1 = (1, -1, -3) m = [tex]\sqrt{1^2 + -1^2 + -3^2}[/tex]
    F1 = (8i - 8j + 4k = 4 N

    I found the distance between Q1 and Q2 using pythagorean formula: 3 meters.
    Then I solved for Q2 using Coulomb's Law:
    Q2 = 13.3 uC

    For the second part, I know that since they ask for the expression of the electric field at the location of Q2, Q2 will not contribute to the value. Beside that, I'm a bit lost.

    Do I need to use the distance between Q1 and Q2, then solve for E=k|q| / r^2? What's |q| in this case?

    Thanks!
     
  2. jcsd
  3. Jul 20, 2010 #2
    The E-field by both charges is not definite at the positions of the charges. I think if the question is given by your teacher, you should seek clarification from him whether he asks for the E-field by both or the E-field contributed by only Q1 (if the latter, ignore Q2 as you only count Q1 in the E-field by Q1). Otherwise, you should ignore the question, as it doesn't have an appropriate answer.
     
  4. Jul 20, 2010 #3
    I'm assuming it's the latter; that I can ignore Q2 and only count Q1. I'm still not sure how to do it though. Is |q| in E=k|q| / r^2 simply the value of Q1?

    Thanks!
     
  5. Jul 20, 2010 #4
    Do you have to count something else besides Q1 when calculating something created by only Q1? :wink:

    EDIT: By the way, the question asks for the E-field which is a vector, not just its magnitude. You may find this formula helpful [tex]\vec{E}=k\frac{q}{r^3}\vec{r}[/tex]
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook