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Zero electric field

  1. Jan 20, 2009 #1
    1. The problem statement, all variables and given/known data

    A charge of -3.02 micro C is located at the origin abd a charge of -2.44 micro C is located along the y axis at 2.6109 m. At what point along the y axis is the electric field zero?

    2. Relevant equations
    E = q/ r^2



    3. The attempt at a solution

    first i determined equations to use for "r" (distance)

    r1 = y2 (2.6109m-y)^2
    r2 =
    I set the electric fields equal to zero

    E2-E1 = o

    k(2.44 micro C)/(6.8168-6.8168y=y^2) - k(3.02 micro C)/y2 =0

    Then i just solved these as a quadratic in the end but the answer under the square root comes out as a negative or as a ridiculously large number. Please guide me through. Much appreciated!
     
  2. jcsd
  3. Jan 20, 2009 #2

    rl.bhat

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    k(2.44 micro C)/(6.8168-6.8168y=y^2) - k(3.02 micro C)/y2 =0

    Wright this in the following form:
    k(2.44 micro C)/(2.6109-y)^2 = k(3.02 micro C)/y^2
    Take square root on both side and find the value of y. While writing down the answer you have mention from which charge you are measuring the distance y.
     
  4. Jan 20, 2009 #3
    why should i take the square root. I think the way i solved it and then used the quadratic formula in the end was much easier...
     
  5. Jan 20, 2009 #4

    rl.bhat

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    (6.8168-6.8168y=y^2)
    Check this one.
     
  6. Jan 20, 2009 #5
    oops sorry that was a typo . its (6.8168-6.8168y+y^2)
     
  7. Jan 20, 2009 #6

    rl.bhat

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    [QUOTE=itryphysics;2041778]oops sorry that was a typo . its (6.8168-6.8168y+y^2)[/QUOTE]

    It is also wrong.
    (2,6109 - y)^2 = ?. Check it.
     
  8. Jan 21, 2009 #7
    argh! i had a brain spasm i guess...messed up on simple multiplication. Thank you so much for pointing that out!
    now i got >> 6.8168-5.2218y+y^2
     
  9. Jan 21, 2009 #8
    so now I think im going to set two equations equal to each other and then solve for y which will be my distance but I dont know which way i should set it up. Please suggest..

    k(2.44 e-6 microC)/ (2.6109-d)^2 = k(3.02 e -6 micro C) / d2

    Is it this way or do i need to swap around the denominators? please explain how you figure this out as well. Thanks
     
  10. Jan 21, 2009 #9

    rl.bhat

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    Yes. Now you can proceed. Still taking square root is more easier.
     
  11. Jan 21, 2009 #10
    so my choice of denominators is correct?
     
  12. Jan 21, 2009 #11

    rl.bhat

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    When you take square root on both side you get
    (2.44 micro C)^1/2/(2.6109-y) = (3.02 micro C)^1/2/y
    Now swap around the denominators and solve for y
     
  13. Jan 21, 2009 #12
    for my answer im getting y= 1.37

    I only have one last try left on my online homework program. Can you please check this with your answer to let me know if I committed any errors.Thanks!!!
     
  14. Jan 21, 2009 #13

    rl.bhat

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    Your answer is correct.
     
  15. Jan 21, 2009 #14
    your help is greatly appreciated!
     
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