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Determine the point at which the electric field is zero

  1. Feb 15, 2004 #1
    Two charges -2.50 C and 6.00 C are 1.00 m apart, determine the point at which the electric field is zero.

    I thought this problem would be easy but I must be doing something ridiculously wrong because I've been stuck on this problem for a about an hour...

    I broke up the problem into E1 & E2 using E = kq/r^2, and replacing r^2 in E1 with x^2 and r^2 in E2 with (x+1.00m)^2 and tried to solve using the quadratic formula without success..

    Any hints?

    thanks in advance :)
     
  2. jcsd
  3. Feb 15, 2004 #2

    drag

    User Avatar
    Science Advisor

    Everything you did is correct.
    You use E1+E2=0, use your xlike you said and
    solve. Try it again.
    (I get about 1.91 m)
     
  4. Feb 15, 2004 #3
    hmmmm..

    It's been a long day so maybe I'm making some stupid mistake, but I'm still unable to come up with anything near the answer.. The book shows (1.82m to the left of the negative charge) as the answer, and you're at ~1.91m, very much closer than my answer.


    Here's what I did:

    E1=KQ/x^2
    E2=KQ/(x+1)^2

    E_net = E1 + E2

    E_net = 6KQ/x^2 - 2.50KQ/(x+1.00)^2

    2.50/(x+1.00m)^2 = 6/x^2

    2.50x^2 = 6(x+1.00m)^2

    2.50x^2 = 6x^2 + 12.00x + 6.00

    6x^2 - 2.50x^2 + 12.00x + 6.00 = 0

    3.50x^2 + 12.00x + 6.00 = 0

    -12.00 +- sqrt (12.00)^2 - 4(3.50)(6.00) / 2(3.50)

    -12.00 +- sqrt (144)-(84) / 7.00

    -12.00 +- 7.75 / 7.00

    = -0.607 and -2.82 which I believe is completely wrong...


    Am I making a stupid calc error? Am I not finishing the problem? or did I just completely approach it wrong?

    Thanks for the hints in advance.. :)
     
  5. Feb 15, 2004 #4

    drag

    User Avatar
    Science Advisor

    Oops... your book is correct I made a calc mistake
    by getting delta = sqrt(100+180) and it's supposed
    to be sqrt(100+140). Which gives you 1.82.

    Anyway, the electric field will be 0 farther (1+x) from the
    6 C than from the -2.5 C (x) so you've simply put
    the distances in reverse in your field strenght equations.

    Live long and prosper.
     
  6. Feb 15, 2004 #5
    Ahhhhhh... I came out with 1.82m now. Thanks!!
     
  7. Feb 16, 2004 #6
    oooride,
    I had that exact problem to work out too. We must be using the same book, "Physics for Scientist and Engineers", 6th edition by Serway and Jewett?

    I got stuck on this one until the prof went over it in class.
    After you posted it here, I went through to see if I can solve it on my own, and I GOT STUCK AGAIN!!!

    Thanks for the review reminder.
     
  8. Feb 16, 2004 #7

    Yup that's the book, except I'm using the fifth edition instead of the sixth.

    The book's title is "Physics for Scientists and Engineers with Modern Physics, Volume 2, by Serway and Beichner".
     
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