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Homework Help: Is the electric field zero?

  1. Apr 21, 2008 #1
    1. The problem statement, all variables and given/known data
    Two point charges are placed along a horizontal axis with the following values and positions: +3.0 �C at x = 0 cm and -7.0 �C at x = 20 cm. At what point along the x axis is the electric field zero?

    2. Relevant equations

    3. The attempt at a solution
    Can I just set Ea = Eb
    Ea = K*A/R^2
    Eb = K*B/R^2
  2. jcsd
  3. Apr 21, 2008 #2
    thats the basic idea.
    but keep in mind that the distances will be different (the R's) and thats what you need to solve for.
    Make 2 variables R1 and R2 (for each distance), how are they related?
  4. Apr 21, 2008 #3
    Oh I thought the distance was just the distance between A and B....but its the distance from point A to the point where the electric field and same for B?
  5. Apr 21, 2008 #4
    correct. this is a little strange, we're using the method of a "test charge" which essentially means a pretend charge (you can call the charge "q" but it should always end up canceling in cases like this). a test charge is a way to "test" the field at a certain point where there isn't actually a charge. if you use the distance R (from A to B), that will give you the field at A and at B, note that the forces will be the same at those points (in accordance with newton's second law), but you are looking for a point inbetween the A and B where the pulls (or pushes) are equal and cancel eachother out.
  6. Apr 21, 2008 #5
    So I just use R?
  7. Apr 21, 2008 #6
    you have to use R1 and R2, where R1 + R2 = R (the total distance)
  8. Apr 21, 2008 #7
    I don't get it lol...I'm sorry...
  9. Apr 21, 2008 #8
    Ea = Eb
    Ea = K*A/R1^2
    Eb = K*B/R2^2
    where R2 = R - R1 (in this case R = 20cm the total distance between A and B).
    You're are trying to find the point C, such that the field (or force) is zero.
    How about that?
  10. Apr 21, 2008 #9
    Oh wow thanks :)....I see now....So I end up with...
    Ea = K*A/R1^2
    Eb = K*B/(R-R1)^2

    A/R1^2 = B/(R-R1)^2
    And I solve for R1??
  11. Apr 21, 2008 #10
    More importantly, does it makes sense why we set it up like that?
  12. Apr 21, 2008 #11
    Yup I completely understand why you set it up like that...thanks!!
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