1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    User Avatar
    Homework Helper

    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


    User Avatar
    Homework Helper

    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


    User Avatar
    Homework Helper

    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


    User Avatar
    Homework Helper

    You found r, but that is the distance along the -x direction. So yes the position would be -r in that direction.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook