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Electric field = 0 at what point on x axis?

  1. Apr 3, 2012 #1
    1. The problem statement, all variables and given/known data

    you have a +3C and a -1C charge on the x axis. the +3C charge is at the origin and the -1C is at x=1. At what point on the positive x axis does E=0?

    2. Relevant equations

    E=kQ/r^2


    3. The attempt at a solution

    So i have

    E=(-1k/r^2)+(3k/(1+r)^2)

    Setting E=0

    1k/r^2=3k/(1+r)^2

    r= sqrt(0.5) = 0.707

    is this correct?
     
  2. jcsd
  3. Apr 3, 2012 #2
    Will your answer be a point in between the two charges, or to the right of each of them?
     
  4. Apr 3, 2012 #3

    ehild

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    What does "r" mean?

    No, r= sqrt(0.5) is not correct. How did you get it?

    ehild
     
  5. Apr 3, 2012 #4
    ah, r=1.707?
     
  6. Apr 3, 2012 #5

    ehild

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    What is r?
     
  7. Apr 4, 2012 #6
    r=the point on the x axis where the E firld is 0

    right?
     
  8. Apr 4, 2012 #7

    ehild

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    You used the expression of the electric field at distance "r" from a point charge. "r" is distance, not a point. What do you substitute for "r" for the 3C and -1C charges?


    ehild
     
  9. Apr 4, 2012 #8
    my bad. r=distance to the right of the -1C charge. x=1.707 is the point on the x axis where E=0.

    right?
     
  10. Apr 4, 2012 #9

    HallsofIvy

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    No, it isn't. Your r is clearly a number, not a point. I suspect you mean it is the x-coordinate of such a point. In that case, the distance to 0 is r but the distance to 1 is |1- r| not 1+ r.
     
  11. Apr 4, 2012 #10

    ehild

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    If so, your equation 1k/r^2=3k/(1+r)^2 is correct, but the solution is wrong. Check it.

    ehild
     
  12. Apr 4, 2012 #11
    WOW i feel STUPID.

    r=-0.366 or 1.366

    Since x is to the right of the -1C charge, x = 2.366 when E=0.

    I hope i am right now.
     
  13. Apr 4, 2012 #12

    ehild

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    It is correct now. Well done!

    ehild
     
  14. Apr 4, 2012 #13
    Thanks man!

    Now, I'm just curious, but if I wanted to find the spot where V=0, I would have

    0=3Q/(1+r) + (-Q/r)

    so r=1/2

    Does that mean V=0 at x=1.5 or V=0 at both x=1.5 and x=0.5?
    I think it is the second one as all potential lines have to form a closed surface.
    Am I right?

    thank you.
     
  15. Apr 4, 2012 #14

    ehild

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    It is less confusing if you calculate with the position of the point on the x axis, that is with its x coordinate.
    If a point charge Q is at position x=a, its contribution to the potential at x is kQ/|x-a|. Note the absolute value in the denominator!

    In this problem, U(x)=3k/|x|-k/|x-1|. U(x)=0 at x=1.5 and x=3/4.

    ehild
     
  16. Apr 5, 2012 #15
    Yes you are right, it is a lot easier to make sense of the answer if i use the point on the x axis.

    Why did you use absolute value?

    I used

    0 = 3k/x + [(-Q)/(x-1)]

    and i get the solution x = 1.5 only

    thanks.
     
  17. Apr 5, 2012 #16

    ehild

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    The term "r" in the denominator of the potential is distance. The distance is positive (or zero). The coordinate can be both positive and negative.
    At distance r from a point charge, the potential is the same either at the right hand side of the charge or at the left hand side.
    Find the potential of the -1 C charge at x=3/4. What do you get without absolute value and what is the right value of the potential?

    ehild
     
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