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Electric potential

  1. May 14, 2006 #1
    Sorry to be a pain again, but I'm struggling with my whole electromagnetic course, that I'm being examined on in a few days!

    Next question I think may just be a problem with arithmetic and algebra, but we'll probably see that its more than that!

    Question says:

    A distribution of three charges is given below. Show that the electric potential at point P on the x-axis, where x>a is given by:
    [tex]V(x) = \frac{x^2 + a^2}{x(x^2 - a^2)}[/tex]​

    [​IMG]

    Now I tried this. My plan was to work out the electric potential from each point and then add them up (linear superposition right?). Having been given that [tex]V(r) = \frac{q}{4\pi \epsilon_{0}r}[/tex] earlier I proceeded to use this.

    Am I correct in thinking that the distance from q1 (the left hand charge) is 2a+x, the distance from q2 (center) is a+x and from q3 is just x?

    Then when I added all these together I ended up still having the q/4pi etc in the fraction as well as another fraction that I got as [tex]\frac{2a-2x}{x(2a+x)}[/tex] instead of the required fraction.

    Does anybody see where I went wrong with this?

    Thanks again guys
     
  2. jcsd
  3. May 14, 2006 #2

    Doc Al

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    Staff: Mentor

    No. x is just the x-coordinate of point P, measured from the origin. The distance between P and q1, for example, will be a + x, not 2a + x; the distance between P and q2 is just x; etc. (At least that's how I interpret the problem statement.)
     
  4. May 14, 2006 #3
    Of course! So for q3 it will be x-a?

    Does the rest of the way I was attempting the question look all right though?
     
  5. May 14, 2006 #4

    Doc Al

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    Right.
    Yes, adding the three potential terms is the way to go.
     
  6. May 14, 2006 #5
    Now I end up with a constant term of [tex]\frac{q}{4\pi\epsilon_{0}}[/tex] outside of a bracket and [tex]\frac{2a^2}{x(x^2 - a^2)}[/tex] within a bracket. I need to get rid of the outside term (q/4pi etc) and I must have gone wrong with my arithmetic.

    Any ideas?
     
  7. May 14, 2006 #6
    Are you sure there is no [itex]\frac{q}{4\pi\epsilon_{0}}[/itex] in the question. Then the question is wrong. Also check your arithmetic.
     
  8. May 14, 2006 #7
    Yep, very sure.

    I've been through a couple of times now, and I end up with the [tex]2a^2[/tex] on the top everytime.

    I'm rapidly losing faith in electromagnetism. My textbook and notes are no help either. :(
     
  9. May 14, 2006 #8

    Doc Al

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    I don't see anything wrong with your solution. (Something's obviously wrong with the book's answer if it leaves out the [itex]\frac{q}{4\pi\epsilon_{0}}[/itex] term!)
     
  10. May 14, 2006 #9
    Have I got my arithmetic wrong?

    I keep trying, but I still end up with [tex]\frac{2a^2}{x(x^2 - a^2)}[/tex]
     
  11. May 14, 2006 #10

    Doc Al

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    As I said before, your answer is perfectly correct. The book is wrong. (Don't waste any more time stewing over this one. Move on to other problems. :smile: )
     
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