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Guass's Law over the x axis

  1. Feb 19, 2013 #1
    1. The problem statement, all variables and given/known data
    Charge is uniformly distributed along the x axis with density ß. Use Gauss' Law to find the electric field it produces, and use this to calculate the work done on a charge Q that moves along the y axis from y = a to y = b.


    2. Relevant equations

    [itex]\phi[/itex]=[itex]\int[/itex][itex]\vec{E}[/itex]*[itex]\hat{n}[/itex]dA

    [itex]\phi[/itex]= [itex]\frac{Q}{\epsilon}[/itex]
    3. The attempt at a solution

    I used a cylinder for my surface since the normal vector will always align with the electrical field. So the first part, the equation ends up
    [itex]\phi[/itex]=E[itex]\int[/itex]dA
    [itex]\phi[/itex]=E(2∏rh)

    (r is the radius from the axis to the edge of the cylinder and h is the length of the cylinder.)

    and if I remember right, Q is the density times the are of enclosure.
    so Q = β(2∏rh)
    I set the two [itex]\phi[/itex] equations equal to each other and get
    [itex]\frac{β}{ε}[/itex]=E

    I don't think that's right though. What did I do wrong?
     
    Last edited: Feb 19, 2013
  2. jcsd
  3. Feb 19, 2013 #2
    Since the charge is along the x-axis, then β is a linear charge density.
    So, to get the charge enclosed, you multiply β by the length h, and not the volume.
     
  4. Feb 19, 2013 #3

    tiny-tim

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    Hi Colts! :smile:
    No, "density" here means the line density (in coulombs per metre, not per metre3)

    So Q = βh. :wink:

    (of course, sometimes "density" means surface density, and occasionally it actually means density! :rolleyes:)

    EDIT: ap123 beat me to it! :biggrin:
     
  5. Feb 19, 2013 #4
    So the electrical field is

    E = [itex]\frac{β}{2πrε}[/itex]
     
    Last edited: Feb 19, 2013
  6. Feb 19, 2013 #5
    Is the last question asking me to integrate E from a to b?
     
  7. Feb 19, 2013 #6

    SammyS

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    Integrate the force, FExt, that would need to be exerted on a charge, Q, to move the charge from a to b . (Actually integrate the work the force does.)
     
  8. Feb 19, 2013 #7
    [itex]\int[/itex][itex]\frac{βxdx}{2πrε}[/itex]

    x is the distance

    does that look right? and the integral would be from a to b
     
  9. Feb 19, 2013 #8

    SammyS

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    Not correct.

    What is the force on a charge Q located on the y-axis , a distance y from the x-axis ?

    You (or some outside agent) must apply what force on Q to move it, at a constant rate, when the charge is located on the y-axis ?

    Added in Edit:

    Another way to do this is to find the potential difference from y = a to y = b .

    To do that, you do integrate -E .
     
    Last edited: Feb 19, 2013
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