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Homework Help: Gauss's Law Question

  1. Oct 7, 2009 #1
    [Solved] Gauss's Law Question

    Well, I've been reading up on Resnick, and have reached the chapter on Gauss' Law.

    He provides an example where he uses Gauss's Law to calculate the E field just outside a positively charged conductor (Arbitrary shape):

    His reasoning was as follows:

    We draw a Gaussian surface in the shape of a small cylinder whose end faces are parallel to the surface of the conductor. Part of the cylinder is just outside the conductor and part is inside. The field is normal to the conductor's surface from the condition of electrostatic equilibrium. (If E had a component parallel to the conductor's surface, the free charges would move along the surface; in such a case, the conductor would not be in equilibrium.) Thus, there is no flux through the curved part of the surface because E is parallel to the surface.
    There is no flux through the flat face of the cylinder inside the conductor because here E=0 (There is no net E field inside a conductor)
    Hence, the net flux through the Gaussian surface is that only through the flat face outside the conductor, where the field is perpendicular to the Gaussian surface.
    Looking at this face, we see that the flux is EA, where E is the electric field just outside the conductor and A is the area of the cylinder's face. Applying Gauss's Law to this surface we obtain:
    [tex]\Phi = EA =\frac{q_{in}}{\epsilon_0}=\frac{\sigma A}{\epsilon_0}[/tex]

    Solving for the field, we obtain:

    (Excerpt from "Fundamentals of Physics")

    Now my question is where exactly we used the fact that we are talking about a surface that's very close to the conductor?
    What keeps me from using a very long cylinder and saying the the contribution of the small surface element [tex]dA[/tex] I was looking at is perpendicular to the surface, and independent of the distance from it?

    With thanks in advance, Anatoli.
    Last edited: Oct 7, 2009
  2. jcsd
  3. Oct 7, 2009 #2


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    Nothing keeps you from using a very long cylinder and in that case the argument in the textbook breaks down because the field far from the conductor is not uniform. What you have here is a derivation based on Gauss' Law that if you come close enough to a charged conductor the electric field will be uniform as if the surface were a plane. It might be intuitively obvious that this is the case, but here is the math to prove it. If you think about it for a moment, the electric field very near the surface of a conductor is uniform in the same sense that the gravitational field near the surface of the Earth is uniform. There is such a thing as Gauss' Law for gravity, you know.
  4. Oct 7, 2009 #3
    That much was obvious, my question is where in the reasoning I used the fact that I'm very close to the conductor.

    I can see how at the limit where I'm approaching the surface, it looks to me as an observer more and more like an infinite plane, but where did I say I was approaching in the above reasoning?

    That's exactly what bothers me. I know that the E field should be of that form when I'm very close, and fall off as [tex]\tfrac{1}{r^2}[/tex] when I'm very far away, instead, this reasoning shows that I can take a very long cylinder and get the same result. I'm looking for the hidden supposition here.
  5. Oct 7, 2009 #4

    Doc Al

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    The "hidden" supposition is just as kuruman stated: Treating E as constant and perpendicular to the surface, which is only true very close to the conducting surface (except in special cases).
  6. Oct 7, 2009 #5
    Oh, I see it now, it is only perpendicular when we are very close to the surface, further out and it propagates radially rather than being perpendicular to the surface, correct?

    The fact that it is a constant when we are close to the surface is a consequence of Gauss's Law and not a supposition in of itself though, isn't it?
  7. Oct 7, 2009 #6

    Doc Al

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    No, it's not a consequence of Gauss's law, it's a supposition we must add in order to use Gauss's law.

    Gauss's law is true regardless of whether E is constant across or perpendicular to any particular Gaussian surface.
  8. Oct 7, 2009 #7
    Oh, thank you very much! I see now. :)
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