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E Field Question

  1. Jan 31, 2008 #1
    1. The problem statement, all variables and given/known data
    Point charge Q1 = 25nC located at P1(4,-2,7) and a charge Q2 = 60 nC be at P2(-3,4,-2). At what point on the y-axis is Ex=0?

    2. Relevant equations
    E= [ Q / (4*pi*E0*r*r) ] ar

    3. The attempt at a solution

    I took:
    ( Q1 / 4*pi*E0*(x*x+y*y+z*z) * (y / sqrt(x*x+y*y+z*z)) ay

    I got -6.51.

    The solution is -6.89.

    Am I way off base or on the right path? Thanks for your help.
  2. jcsd
  3. Jan 31, 2008 #2
    It appears I might have been way off base. Reworking a different thought...call off the Help Department (for now). :)

    Report back in a few...
  4. Jan 31, 2008 #3
    Ok. I'm stuck again...
    Basically we want Ex=0 at some point (P3) on the y-axis. So the point P3 (0,y,0).

    Taking the two points to get the total E field. ET = E13 + E23.

    I've got:

    E13 = 25nC [(0-4)ax + (y+2)ay + (0-7)az ] / (4*pi* 8.854e-12 [(-4)^2 + (y+2)^2 + (-7)^2]^3/2)

    E23 = 60nC [(0-(-3)ax + (y-4)ay + (0+2)az ] / (4*pi* 8.854e-12 [(3)^2 + (y-4)^2 + (2)^2]^3/2)

    Kind of stuck at this point...any help? Thanks.
    Last edited: Jan 31, 2008
  5. Feb 22, 2008 #4
    I saw throught your work very quickly. Here's what I have to offer:

    Now since you want Ex= 0 equate the x components of E13 and E23 together and take their magnitude. Only when they are EQUAL and OPPOSITE in magnitude and direction respectively, will you get Ex= 0.
  6. Feb 22, 2008 #5
    Yes, that works! You can do it in two ways. From the above post, you can either calculate the value for y, or you can plug in your answer y= -6.89 from the solutions manual that you have and find out that both the magnitudes are the same.


    25nC* -4
    ----------------------------- = -0.11927*10^-9 when y= -6.89
    [16+ (y+2)^2 + 49 ]^3/2


    60nC* 3
    ----------------------------- = -0.11924*10^-9 when y= -6.89
    [9+ (y-4)^2 + 4 ]^3/2

    I think that's pretty close enough!

    I hope that helped.
  7. Feb 22, 2008 #6
    Yes, it did. Thanks for your help.
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