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Homework Help: Finding net electric field at different locations

  1. Sep 16, 2008 #1
    1. The problem statement, all variables and given/known data
    Two charges are placed on the x axis. One charge (q1 = +16.5 µC) is at x1 = +3.0 cm and the other (q2 = -19 µC) is at x2 = +9.0 cm. Find the net electric field (magnitude and direction) at the following locations.

    (a) x = 0 cm
    magnitude= ? N/C
    direction= ?


    (b) x = +6.0 cm
    magnitude= ? N/C
    direction= ?



    I set E1=E2

    [K(+16.5*10^-6)]/X^2 = [K(-19*10^-6)]/(X+.06)^2

    After canceling everything out I get 16.5(X+.06)^2 = 19X^2

    Then solving for X I get .8219 and -.0289

    I am not sure what I am doing wrong. The (10^-6) should cancel out right?
     
  2. jcsd
  3. Sep 16, 2008 #2

    LowlyPion

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    First of all, welcome to PF.

    But as to setting E1=E2 ... Why?

    The E field at any point is

    [tex]\sum \vec{E}[/tex]

    Perhaps a better approach is determining the E field contribution from each charge and adding them together remembering of course that you are adding E field vectors?
     
  4. Sep 16, 2008 #3
    So, It would be "[K(+16.5*10^-6)]/X^2 + [K(-19*10^-6)]/(X+.06)^2" instead?

    I was following another example I found and now I am very confused.
     
  5. Sep 16, 2008 #4

    LowlyPion

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    Perhaps it would be less confusing if you dropped X from your equations. And then calculated each separately and added?

    What is crucial is the distance from the charge. That and Coulomb's constant and the charge determine the E field contribution from that point.

    So in the first one the E-field for the first charge at x= .03 m (dimensions in meters for the constant to be valid) resolves as:

    [tex]\vec{E_1} = \frac{k*q_1}{r^2} = \frac{k*16*10^{-6}}{.009}[/tex]

    Then calculate the contribution for the negative charge that is at .09m.

    Those two together determine the field at x=0.

    [tex]\vec{E} = \vec{E_1} + \vec{E_2}[/tex]
     
  6. Sep 16, 2008 #5
    When I do that I get -5111111.111

    E= [(9x10^9)(16x10^-6)]/(.009) + [(9x10^9)(-19x10^-6)]/(.0081)

    What am I doing wrong?
     
  7. Sep 16, 2008 #6

    LowlyPion

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    3 cm2 is .0009 not .009 as I originally typed. My mistake.

    that makes it 16*107 - 2.1*107 = 13.9*107
     
  8. Sep 16, 2008 #7
    So, for part B I would just change the radius value?
     
  9. Sep 16, 2008 #8

    LowlyPion

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    Pretty much that looks like it.

    Units are N/C - Newtons/Coulomb
     
  10. Sep 16, 2008 #9
    Thank you so much. I have a hard time understanding Physics and this really helped.
     
  11. Sep 16, 2008 #10

    LowlyPion

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    Great. Good Luck then.
     
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