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Homework Help: Electric Field Question

  1. Jan 29, 2008 #1
    1. The problem statement, all variables and given/known data

    http://img184.imageshack.us/img184/8125/chapter23number13qh8.png [Broken]

    2. Relevant equations

    [tex] E = {k}_{e} \frac{q} {r^2}

    3. The attempt at a solution

    I figured that the electric field vector from the negative point charge would have to cancel out the electric field vector from the positive point charge so I said that 0 = E from the negative + E from the positive. I substituted values in and made the distance for the positive one r and the distance for the negative one (1-r). When I solved this for r, I get an answer like 1.03m, but the book says the answer is 1.82m to the left of the negative charge.

    Even if I did the math wrong, what I am really wondering about it is the concept here. To me it seems like there should be 4 points where the field is 0. 1. At infinity, which the book mentions. 2. Somewhere far to the left of the negative charge, which I guess is the answer they put in the back. 3. Just slightly to the left of the negative charge. 4. And just slightly right of the negative charge.

    My thinking on 3 and 4 is that E will get bigger and bigger as you get closer to the point (r->0, so E goes to infinity, right?). So, E will be large and positive close to the (+) charge, but as you go left to the (-) charge it will decrease to 0, and then become more negative as it reaches the (-) charge. Once it passes this, it will get bigger (from -infinity) until it equalizes with the (+) charge. Then, when you are rather far away it will be 0 again.
    Last edited by a moderator: May 3, 2017
  2. jcsd
  3. Jan 30, 2008 #2
    Re 3: How can the field add up to zero, both "far to the left"(not infinity) and "slightly to the left" of the negative charge? Only one can be true.

    And as for 4, what are the directions of the electric fields at a point between the charges (along the line joining them)?
  4. Jan 30, 2008 #3
    Ohhhh I get it now! Thanks for the help!
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