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Gauss's Law question

  1. Oct 19, 2008 #1

    TFM

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    1. The problem statement, all variables and given/known data

    A long thin wire carries a charge [tex]\lambda[/tex] per unit length

    a) Use Gauss's Law on a cylinder to find magnitude of the E-Field at a distance from the centre of the wire

    b) Show that the electric field components (using cartesian coordinates) are of the form:

    [tex] E_x = \frac{A_0x}{r^2}

    E_y = \frac{A_0y}{r^2}

    A_0 = \frac{\lambda}{2\pi\epsilon_0}

    [/tex]

    2. Relevant equations

    Gauss Law:

    [tex]\int_sE.dA=\frac{q}{\epsilon_0}[/tex]

    3. The attempt at a solution

    I have the asnwer for a) to be

    [tex] E = \frac{q}{\pi r^2 \epsilon_0} [/tex]

    But for B:

    [tex] E_x = \frac{\lambda x}{2\pi epsilon_0} [/tex]

    But I get:

    [tex] q = \lambda x [/tex]

    giving:

    [tex] E = \frac{q
    \lambda x}{\pi r^2 \epsilon_0} [/tex]

    a factor of a half out.

    Any ideas where I could have gone wrong?

    TFM
     
  2. jcsd
  3. Oct 19, 2008 #2

    TFM

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    Latex appears to be not working :bugeye:

    1. The problem statement, all variables and given/known data

    A long thin wire carries a charge lambda per unit length


    a) Use Gauss's Law on a cylinder to find magnitude of the E-Field at a distance from the centre of the wire

    b) Show that the electric field components (using cartesian coordinates) are of the form:

    E_x = \frac{A_0x}{r^2}

    E_y = \frac{A_0y}{r^2}

    A_0 = \frac{\lambda}{2\pi\epsilon_0}

    2. Relevant equations

    Gauss Law:

    \int_sE.dA=\frac{q}{\epsilon_0}

    3. The attempt at a solution

    I have the asnwer for a) to be

    E = q/(pi r^2 \epsilon_0)

    But for B:

    E_x = lambda x/2\pi epsilon_0

    But I get:

    q = \lambda x

    giving:

    E = \frac{\lambda x/(pi r^2 \epsilon_0)

    a factor of a half out.

    Any ideas where I could have gone wrong?

    TFM

    Edit: I don't think the non latex version helps!!!
     
  4. Oct 19, 2008 #3

    Doc Al

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    Staff: Mentor

    Yeah, it's being worked on. :cry:

    Show how you got that.
     
  5. Oct 20, 2008 #4

    TFM

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    Well Gauss's law:

    [tex]
    \int_sE.dA=\frac{q}{\epsilon_0}
    [/tex]

    intergral E.dA = q/epsilon-0

    the area is a circle, pi r squared, so I get

    E*pi r squared = q/epsilon-0

    divide by pi r squared gives:

    E = q/(epsilon-0 pi r squared)

    [tex] E = \frac{q}{\epsilon_0 \pi r^2} [/tex]

    TFM
     
  6. Oct 20, 2008 #5
    You used the area of a circle, which is incorrect: you need the surface area of a cylinder. That has something to do with a circle indeed, but not with its area! Also, you have to calculate what the total charge inside the cylinder is, you can't just write q=\lambda x (wrong, too!)
     
  7. Oct 20, 2008 #6

    TFM

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    So far, I now have:

    E(2 pi r h + 2 pi r squared) = q/epsilon-0

    [tex] E (2*\pi*rh + 2 \pi * r^2 = \frac{q}{\epsilon_0}) [/tex]

    so E = q/(epsilon-0 [2 pi r h + 2 pi r squared])

    How does this look now?

    TFM
     
  8. Oct 20, 2008 #7
    You're halfway there:

    The flux of E goes only through the 'side surface", not the top and bottom surfaces.
    That is, the direction of E is perpendicular to the normal vectors pointing out of the top and bottom surfaces.
     
  9. Oct 20, 2008 #8

    TFM

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    I see, so it will be

    E(2 pi r h + 2 pi r squared) = q/epsilon-0

    [tex] E (2*\pi*rh) = \frac{q}{\epsilon_0}) [/tex]

    so E = q/(epsilon-0 2 pi r h)

    as the flux isn't coming out the two circles at the top and bottom

    TFM[/QUOTE]
     
  10. Oct 20, 2008 #9
    The left-hand side of your first equation still contains two contributions, flux through the sides, and flux through the top/bottom. Get rid of the second part.

    Your final answer looks much better now, but what is q?

    Also, insert unit vectors for the direction of the E-field.
     
  11. Oct 20, 2008 #10

    TFM

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    I see, so it will be

    E(2 pi r h) = q/epsilon-0

    [tex] E (2*\pi*rh) = \frac{q}{\epsilon_0}) [/tex]

    so E = q/(epsilon-0 2 pi r h)

    as the flux isn't coming out the two circles at the top and bottom

    Is lamda not the charge density? If it is, why isn't it lambda * d <- the length of the wire?

    TFM
     
  12. Oct 20, 2008 #11
    You have an "h" in your answer: what's that quantity? Is q related to h?
     
  13. Oct 20, 2008 #12

    TFM

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    h is the length of the wire, and the charge density is lambda - charge ;per unit length, so I'm guessing it is related.
     
  14. Oct 20, 2008 #13
  15. Oct 20, 2008 #14

    TFM

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    I'd say lambda*d = q, for the wire

    TFM
     
  16. Oct 20, 2008 #15
    Hey TFM,

    That answer looks good to me.
     
  17. Oct 20, 2008 #16

    TFM

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    So

    E = (lambda d)/(epsilon-0 2 pi r h)

    ?

    TFM
     
  18. Oct 20, 2008 #17
    Almost. If you make your Gaussian surface as long as the wire, what will happen to h?

    Edit: Sorry I meant h not d.
     
    Last edited: Oct 20, 2008
  19. Oct 20, 2008 #18

    TFM

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    h will be the same length as the wire, owuld it mnake the charge lambda?

    TFM
     
  20. Oct 20, 2008 #19
    Lambda would still be the charge per unit length, but h would drop out of the equation, leaving just lambda/(2*pi*r*epsilon0), which is the electric field at a distance r due to an infinitely long wire. In this case, since it is only a long wire, we would have to stipulate r<<h.

    edit I meant r<<d
     
    Last edited: Oct 21, 2008
  21. Oct 20, 2008 #20

    TFM

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    So:

    E = lambda/(2*pi*r*epsilon-0)

    but that doesn't seem to help with the next part, which is showing that the x-component is:

    E_x = (lambda x)/2*pi*epsilon-0 r^2

    and the y-component

    E_y = (lambda y)/2*pi*epsilon-0 r^2

    ???

    Daft-Vader
     
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