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Homework Help: Electric field due to two charges PLEASE HELP!

  1. Sep 30, 2004 #1
    Find the point (or points) where the electric field due to two charges -5q and +2q is zero and separated by a distance a metres.

    -5q ------------------- +2q

    WEll first of i know that there is a point that is along the axis on the right side closer to 2q but farther from the negative.

    But i'm wondering if there is a point somewhere else is space (other than infinity) where the electric field is zero.

    Since the electric field due to the negative is pointing towards it and positive points outward i wonder if there is a point where the two vectors cancel out anywhere else is space.

    Am i right in assuming that there is no other point besides the point on the axis??
  2. jcsd
  3. Sep 30, 2004 #2
    i mean the axis upon which they are mounted

    so there is a point along this axis where E = 0 , however, no other point exists?
  4. Sep 30, 2004 #3
    but how about a point t such that 2kq/t^2 - 5kq/(a+t)^2 = 0 and i can solve for t and it is a positive number
  5. Sep 30, 2004 #4


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    Hmm...I'm taking electromagnetism right now, so I wanted to make sure I understood why he is correct. Call the left charge q1, the right q2:

    [tex] \vec{E}_1 = \frac{1}{4\pi\epsilon_0}\frac{q_1}{r_1^2}\hat{r}_1 [/tex]

    [tex] \vec{E}_2 = \frac{1}{4\pi\epsilon_0}\frac{q_2}{r_2^2}\hat{r}_2 [/tex]

    r_1 and r_2 are the magnitudes of the separation between the field point and the first and second source points (respectively)

    [tex] \hat{r_1}\ \ \text{and}\ \ \hat{r_2} [/tex]

    are unit vectors pointing in the direction from each source point to the field point. By the principle of superposition, the net electric field is the vector sum of the two, and this sum can only be zero for some point at which:

    [tex] \hat{r_1} = -\hat{r_2} [/tex]

    and this is only true along the line of action between the two charges. For any other point, the vector sum will yield some resultant electric field vector.

    EDIT: Never mind, I'm wrong. I was thinking about two like charges. I just looked at vsages diagram, and it seems that [itex] \hat{r_1} = \hat{r_2} [/itex] along the line of action between them i.e. the fields due to each one always point in the same direction. Thanks for clearing this up.
    Last edited: Sep 30, 2004
  6. Sep 30, 2004 #5
    So there's no point?

    You're right both the fields almost reinforce each other with direction, one points away one towards it, so this maenas there exists no point???!
  7. Sep 30, 2004 #6

    thank you very much

    i am in your debt
  8. Sep 30, 2004 #7


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    Of course, there is a point (other than at infinity) where the two E fields cancel out to give a total E field equal to zero. It is to the right of the 2q (as I think you pointed out in your first post). I am not sure what vsage meant (maybe he/she was just focusing on points between th echarges?? I don't know).

    But other than that point (which you can find by setting the two magnitudes equal), there is no other point with E_total=0 (other than at infinity)

  9. Sep 30, 2004 #8
    ok im wondering who's the superior authority on this

    that's what i thought in the first place
    Last edited: Sep 30, 2004
  10. Sep 30, 2004 #9


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    The point located at a distance [itex] {{\sqrt 2} \over {\sqrt 5} - {\sqrt 2}} a \approx 1.72 a [/itex] to the right of the 2q has a total E field equal to zero.

  11. Sep 30, 2004 #10
    but what about the direction

    you would put the two equal to each other

    2kq / t^2 = -5kq / (a+t)^2

    or would they bne the same signs?
  12. Sep 30, 2004 #11


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    I am glad you read my post because I was worried you would be left with the wrong idea!! (btw, I give the exact position in another posting).

    As fo the superior authority, I can give you my credentials (PhD in phys, 8 years of college level teaching) but in the end, it must be decided by physics and maths. If you pick a point to the right of the 2q, the two E fields point in opposite directions. And because you are closer to the weaker charge (in absolute value), there is necessarily a point where the two magnitudes will be equal. Call "x" the distance between that point and the 2q charge. Then you only need to solve 2q/x^2 = 5q/(x+a)^2 (I used the absolute values of the charges since I am imposing the two magnitudes to be equal). Solving for x gives the answer I presented in my other post.

  13. Sep 30, 2004 #12

    i'm convinced, thank you
  14. Sep 30, 2004 #13
    Hrm I am going to bring this up at my next physics lecture. I'm certainly open to ideas but I just hadn't thought of it as possible before. Sorry stunner! I would go with Nrq definitely. I am a first year college student (engineering major) so Nrq has quite a bit of education on me.
  15. Sep 30, 2004 #14
    no problem
  16. Sep 30, 2004 #15


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    This type of problem is better done in two steps. In the first step, you worry about the directions only: you look for points where the two E field produced by the two charges are in opposite directions. In this case, it's easy to see that any point to the left of the -5q or to the right of the +2q satisfy this (if the charges had been of equal signs, you would have picked a point somewhere between the charges).

    Now, you consider the magnitudes. The vectors must not only be pointing in opposite directions, they must have the same magnitude. This rules out the points on the left of the 5q (because these points are always closer to the stronger charge (in absolute value) so there is no way for the two magnitudes to be equal). That leaves the points to the right of the 2q.

    Now, you impose that the two *magnitudes* be equal, which means you impose

    [itex] { |-5 q| \over (x+a)^2 } = { |2 q| \over x^2} [/itex]

    where my vertical bars denote the absolute values. And then you solve.

    Basically, what we did is to solve the condition [itex] \vec E_1 + \vec E_2 = 0 [/itex] or [itex] \vec E_1 = - \vec E_2 [/itex] which means teh two vectors must be opposite and of equal magnitude. We took care of the directions first and the when came the time of finding the exact position, we work with the magnitudes.

    I hope this clarifies things for you.

  17. Sep 30, 2004 #16
    While we're on this topic can you assist me in something similar to this

    ill post in a new thread though
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