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Homework Help: Where electric field is zero

  1. Sep 27, 2008 #1
    1. The problem statement, all variables and given/known data

    an electron (charge -e) is at the origin, and a particle of charge +5e is at x = 17nm, find a point (x = ...nm) where the electric field is zero.

    2. Relevant equations

    e = 1.6*10^-19 coulombs
    electric field E = kq/r^2 where k=9*10^9, q is charge, r is distance

    3. The attempt at a solution

    kq_1/r^2 = kq_2/(17-r)^2

    k(-e)/r^2 = k(+5e)/(17-r)^2

    k(-e)(17-r)^2 = k(+5e)(r^2)

    k(-e)(28 - 34r + r^2) = k(+5e)(r^2)

    (9*10^9)(-1.6*10^-19)(28 -34r + r^2) = (9*10^-9)(8*10^-19)(r^2)

    -1.4*10^-9[28 - 34r -r^2] = (9*10^-9)(8*10^-19)(r^2)

    (-4.032*10^-8) + (4.896*10^-8)r - (1.6*10^-19)r^2 = (7.2*10^-9)r^2

    (-4.032*10^-8) + (4.896*10^-8)r - (7.2*10^-9)r^2 = 0

    quadratic formula

    r = [-4.032*10^-8 +/- sqrt[(4.896*10^-8)^2 - (4)(-7.2*10^-9)(-4.032*10^-8)] ] / (2)(-7.2*10^-9)

    r = [-4.89*10^-8 +/- sqrt(1.237*10^-15)] / -1.44*10^-8

    r = 0.957nm or 5.83 nm

    are my calculations correct? which one is the correct answer?
     
  2. jcsd
  3. Sep 27, 2008 #2

    LowlyPion

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    Perhaps you want to examine your equation a little more carefully.
    Are you sure you want to find a point between two charges of opposite sign where the E-field would be 0?
     
  4. Sep 28, 2008 #3
    are you saying the it should be (17+r) instead of (17-r) or asking conceptually? conceptually, if the point is placed a huge distance away, then the electric field would essentially be zero
     
  5. Sep 28, 2008 #4

    LowlyPion

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    Is there any point between a positive and negative charge that the field can ever be 0?
     
  6. Sep 28, 2008 #5
    no, because the field woulf come out of the positive and flow into the negative charge, correct?

    should it be (17+r) not (17-r), though?
     
  7. Sep 29, 2008 #6
    when i made it 17+r the quadratic i would get would be an imaginary number, what am i doing wrong?
     
  8. Sep 29, 2008 #7

    LowlyPion

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    It must be 17+r from the larger charge.

    Then you can say e/r2 =5e/(17+r)2

    Take the sqrt of both sides and rearrange:
    (17+r) = (5)1/2(r)

    r = 17/(51/2 - 1)
     
  9. Sep 29, 2008 #8
    oh, that simple i got 13.75, but actually it was -13.75
     
  10. Sep 29, 2008 #9

    LowlyPion

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    You found r, but that is the distance along the -x direction. So yes the position would be -r in that direction.
     
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