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Electric Field and Distance From Particle

  1. Jan 12, 2007 #1
    1. The problem statement, all variables and given/known data
    Two particles with positive charges q_1= 0.540 nC and q_2= 8.20 nC are separated by a distance of 1.30 m.


    Along the line connecting the two charges, at what distance from the charge q_1 is the total electric field from the two charges zero?




    2. Relevant equations
    E = k*q/r^2

    d = [-b +/ sqrt(b^2 - 4ac)]/[2a]


    3. The attempt at a solution

    I tried drawing a picture:

    q_1 (0.540 nC) O____>____.____<____O q_2 (8.20 nC)

    O = charged particle
    . = field point
    The arrows show the direction of the electric fields. (Correct????)

    x+y = 1.30 m
    y = 1.30 m – x

    E = (k*q_1)/(x^2) – (k*q_2)/(y^2)
    0 = (k*q_1)/(x^2) – (k*q_2)/(y^2)
    0 = (k*q_1)/(x^2) – (k*q_2)/(1.30-x)^2

    0 = (k*q_1)/(x^2) – (k*q_2)/(x^2-2.60x+1.69)

    (k*q_2)/(x^2-2.60x+1.69) = (k*q_1)/(x^2)

    Simplifying:

    x^2*(k*q_2) – k*q_1(x^2-2.60x+1.69) = 0

    x^2*(k*q_2) – (k*q_1*x^2) + (2.60*x*k*q_1) – 1.69*k*q_1 = 0

    or

    68.848x^2 + k*q_1*(2.60x –1.69) = 0

    Using quadratic equation, x = 0.265 m or –0.449 m

    Distance from q_1 = 0.265 m????????

    Thanks.
     
  2. jcsd
  3. Jan 12, 2007 #2

    Gokul43201

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    Staff Emeritus
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    Gold Member

    That's correct.

    Also, take a look at this (it's doing the same thing, essentially):

    [tex]q_1/r_1^2 = q_2/r_2^2 \implies r_1/r_2 = \sqrt{q_1/q_2} = 0.257 [/tex]

    [tex]\implies r_1+r_2 = 1.257r_2=1.3m \implies r_2=1.035m~,~~r_1=0.265m[/tex]
     
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