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Gauss's Law with a Cylindrical Shell

  1. Feb 18, 2009 #1
    1. The problem statement, all variables and given/known data
    A cylindrical shell of radius 7.00 cm and length 24.0 cm has its charge uniformly distributed on its curved surface. The magnitude of the electric field at a point 19.0cm radially outward from its axis (measured from the midpoint of the shell) is 36.0 kN/C. Find (a) the net charge on the shell and (b) the electric field at a point 4.00 cm from the axis, measured radially outward from the midpoint of the shell.


    2. Relevant equations
    Gauss's law, Coulomb's law, [tex]\rho=Q/v[/tex], [tex]E=F_e/q[/tex]
    [tex]\Phi_e[/tex]=EA (if the vectors E and A are in the same direction...right?)

    3. The attempt at a solution
    My first issue is whether Gauss's law can even be applied here. Is this a closed surface? Is there anything I can do to make it...closed? I know that if i was to make a gaussian surface, it would be to satisfy at least one of the four objectives to simplify [tex]\int[/tex]E[tex]\bullet[/tex]dA where the integrand is the dot product of vectors E and dA (sorry my latex skills aren't good, I'm still figuring it out)
    1. The value of the electric flux is constant over the surface by symmetry.
    2. The dot product of vecotrs E and dA can be expressed as EdA because the vectors E and dA are parallel.
    3. The dot product of vecotrs E and dA is zero because they are perpendicular to eachother.
    4. The electric field is zero.

    So my plan was to make two surfaces, one where R(the radius of the shell) is < r(the radius of the gaussian cylinder) and one where R>r on the same axis as the shell. This would satisfy 1 and 2 I think.

    But then I started thinking that because this shell has a finite length and other dimensions, that E is not going to be the same everywhere, like at the ends, the direction of E isn't just going to be simply radially outward.
    However, if we just choose dA to be a piece away from the ends on the curved surface, we should be all set right?

    So my main questions would be: is this a closed surface? and would my plan be sufficient to figure out the flux?

    Also, what do they mean by net charge on the surface? They're talking about flux right? Am I on the right track? I don't actually know what I would do when I got the flux from my two cylinders.

    That would be done by just taking [tex]\int[/tex]E[tex]\bullet[/tex]dA=E2piRL
    and then we would take the E to beeee umm bah I'm confused! Maybe I just need some general pointers
     
  2. jcsd
  3. Feb 18, 2009 #2

    Tom Mattson

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    Gold Member

    Not quite. The problem says that the charge is distributed over the surface (not throughout the volume), so instead of [itex]\rho=Q/v[/itex] use [itex]\sigma=Q/A[/itex].

    In this case, yes. And [itex]\Phi=\vec{E}\cdot\vec{A}[/itex] (a dot product).

    Sure, just define it to be closed. Gaussian surfaces don't exist, they are mathematical constructs.

    As for your "plan", it seems way too complicated to me. Just define a cylindrical Gaussian surface with height 24 cm and radius 19 cm that encloses the shell and that has the same axis as the shell. Then apply Gauss' law to it. You should have the total charge instantly.

    No, charge isn't flux. Gauss' law states the following.

    [tex]\oint_A\vec{E}\cdot d\vec{A}=\frac{Q_{enc}}{\epsilon_0}[/itex]

    The integral on the right is the electric flux through the surface, and the [itex]Q_{enc}[/itex] on the left is the charge enclosed by the surface.
     
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