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Gaussian surface, and uniform charge density

  1. Feb 8, 2005 #1
    28. The problem is stated as:
    A long, thin wire carrying 5.6nC/m runs down the center of a long, thin-walled, hollow pipe with radius 1.0cm carrying -4.2nC/m spread uniformly over its surface. (a)Find the electric field 0.50 cm from the wire. (b)Find the electric field 1.5 cm from the wire.

    *Ok, I'm not exactly sure how to go at this problem. I'm really confused with gaussian surfaces, and the equations in general.

    The following is my attempt at the 0.5cm away from the wire.

    Enclose the wire in a guassian cylinder with length [itex]l[/itex].
    [tex]
    \lambda (q_{wire})=5.6 \times 10^{-9} \frac{C}{m}
    [/tex]

    Cylinder properties:
    [tex]
    r=0.0005m
    [/tex]
    [tex]
    dA=2\pi=2\pi(0.0005m)
    [/tex]

    Now to the electric field:
    [tex]
    E = \frac{\lambda l}{dA* \epsilon_{0}) l}=\frac{\lambda}{dA \epsilon_{0}}
    [/tex]
    [tex]
    = \frac{5.6 \times 10^{-9} \frac{C}{m}}
    {2 \pi (0.0005m)\frac{1}{4 \pi k}} = \frac{(5.6 \times 10^{-9})(C)(4 \pi k)}{2 \pi (0.0005m)m} = \frac{2.24 \times 10^{-8} \pi C k}{0.001 \pi m^{2}} = \frac{2.2 \times 10^{-5} C (9.0 \times 10^{9} N m^{2}))}{m^{2} C^{2}}
    [/tex]
    [tex]
    = 1.98 \times 10^{5} \frac{N}{C}
    [/tex]

    Is this right? Can I just assume that since I am "enclosing" the surface and not worrying about the ends that the electric field is equal to what I got? Or, do I need to worry about the surface charge of the "long hollow thin walled pipe" ?

    What about the electric field 1.5cm away from the wire?

    I'm thinking this is wrong, and that I need to worry about the pipe, due to the superposition principal... but ... ahh hell, I'm just confused.
    thanks in advance.
     
  2. jcsd
  3. Feb 8, 2005 #2

    Doc Al

    User Avatar

    Staff: Mentor

    I did not check your arithmetic, but the electric field at any given distance r depends only on the charge within the guassian cylinder at r. The ends contribute no flux (the field is radial) and thus the field is:
    [tex]E = \frac{\lambda}{2\pi r \epsilon_0}[/tex]

    For r = 0.005 m (not 0.0005) [itex]\lambda[/itex] is just that of the wire; For r = 0.015 m [itex]\lambda[/itex] is the combined charge from the pipe and the wire.
     
  4. Feb 8, 2005 #3
    For these problems, theres usually a pattern to follow,

    First figure out the amount of charge enclosed by the Gaussian surface.

    In the first problem, the G-surface (cylinder) only encloses charge from the wire, so the charge enclosed

    [tex] Q_{encl} = l \lambda_w [/tex]

    where l is the lenght of the cylinder, and [itex] \lambda_w [/itex] is the linear charge density of the wire.

    Gausses law states the the net FLUX through the surface equals the charge enclosed divided by epsilon nought. that is

    [tex] \oint \vec{E} \cdot \vec{dA} = \frac{Q_{encl}}{\epsilon_0}[/tex]

    The flux is the dot product of E and dA, so were only concerned with the component of E that is perpendicular to the surface.
    Normally in a beginning class, the problem is symmetrical so the E can be taken out of the integral and all you need to calcuate is the surface area (which makes things easy, no messy surface integrals to evaluate :smile: )

    In this situation, every point on the "barrel" part of the cylinder (everything but the the sides) are equi-distant from the wire, and the electric field is perpendicular at every point, there for the value of E is constant and can be taken out of the integral.

    on the sides of the cylinder however, the electric field is parallel to the surface. Therefore the sides contribute no flux, and can be ignored.

    so the Gauss's law equation for this example looks like

    [tex]E \oint \vec{dA} = \frac{ \lambda_w l}{\epsilon_0} [/tex]

    where the surface integral equals the area over the barrel part of the cylinder, and not the sides. from geometry you know the area of this part of a cyllinder is [tex] 2 \pi r l [/tex]
    where r is the radius and l is the height of the cylinder.

    thus, the equation becomes

    [tex] E ( 2 \pi r l ) = \frac{\lambda_w l}{\epsilon_0} [/tex]

    and solving for E you get,

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

    See if you can do the second part. Keep in mind about how much charge is enclosed by your surface.
     
  5. Feb 8, 2005 #4
    You guys ROCK. Thank you so much, both helped me very much.

    grr... I don't know why I put 0.0005m, I actually did that throughout my entire calculation too. :(

    thanks again
     
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