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Gauss’ Law – Don’t see how it proves that E field is zero anywhere inside a hollow sp

  1. Jan 7, 2012 #1
    While I do see how this makes sense using Newton’s Shell method, I don’t see how Gauss’ Law of Flux for a closed surface proves the same thing.

    Both Gauss’ Law of Flux and Newton’s Shell method make perfect sense to me in showing that when dealing with a point outside the hollow conducting sphere, the sphere can be treated as a point charge equal to the sum of the surface charges on the hollow sphere (in Newton’s case it would be the sum of the masses). Since Gauss’ Law (as I understand it – maybe this is where my problem lies) is the sum of differentials of the component of the local E field normal to a differential area on the closed surface dA, multiplied by that differential area, yielding units of Newtons/Coulomb X metres squared; if applying this law to a spherical surface about a central charge Q, then the integral is simple, since the distance from any point on the surface to Q is R and we get that the flux for the sphere works out to be the Charge inside divided by epsilon zero. Fine makes, sense to me.

    But in watching an MIT lecture the professor then applies this law to a hollow sphere and a point inside it. He draws a dashed closed surface inside the sphere upon which the point in question sits and says, well there’s no Q inside that imaginary control surface and thus there is no flux through the surface, i.e. no E field inside the surface. Voila. Voila? What in god’s name is he talking about? Is this some kind of reverse logic? I know there’s no charge inside the surface but there is charge outside the surface which could cause a flux through the surface but for overall field cancellation which Newton’s method actually does prove in a logic way. The MIT prof. professes implies that Guass’ law bypasses the need for Newton’s proof. I just don’t see how; I’d love to, but I just don’t see it. Your thoughts?

    Sincerely,
    Ben.
     
  2. jcsd
  3. Jan 7, 2012 #2
    Re: Gauss’ Law – Don’t see how it proves that E field is zero anywhere inside a hollo

    An intuitive way to think of it if you are interested.

    Think of an electric field line created by a charge entering the surface of a sphere or any surface which encloses some volume, is there a way for it to enter the surface and not leave it?
     
  4. Jan 7, 2012 #3
    Re: Gauss’ Law – Don’t see how it proves that E field is zero anywhere inside a hollo

    Gauss' Law let you calculate the flux through a surface without calculating the electric field at each point on the surface, it just cares about the charge inside the surface.
    Imagine to have a spherical conductor and a point charge. In this case the electric field in between the sphere and the point charge (assuming the sphere has a certain charge on it) is created by both the electric field of the sphere and that of the point charge. Then you want to calculate the flux across a certain closed surface that contains the sphere. To do so you have two ways: you explicitly evaluate an expression for the electric field that allows you to calculate ∫EdA (and this E is given by the superimposition of the electric field given by the point charge with that given by the sphere) or you calculate the magnitude of the charge inside the surface so to do q/ε0. Both results represent the flux across our surface.
    Take the case of a sphere with a hollow inside, it's clear that inside there is no charge. If you want to calculate the flux of the electric field inside the sphere you can take a surface inside the hollow and see the charges.
     
  5. Jan 7, 2012 #4

    SammyS

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    Re: Gauss’ Law – Don’t see how it proves that E field is zero anywhere inside a hollo

    Hello physmeb . Welcome to PF !


    The way I look at this is as follows.

    From the very nature of conductors, it should be clear that (under static conditions)):
    1. The only place excess charge can reside is on a surface.

    2. The electric field within conducting material is zero.​
    So, whenever you draw a Gaussian surface completely with a conductor, the electric flux passing through that surface is zero, independent of Gauss's Law. Gauss's Law then comes into play and tells you that the net charge within that surface is zero. For instance; If there's a cavity within a conductor which contains an isolated charge, Q, then there is a charge of -Q distributed on the surface of that cavity.

    I hope that helps.
     
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