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Gauss' Law and it's meaning

  1. Jul 20, 2010 #1

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    Hello all. I have been trying to learn some Physics in my spare time and I came across Gauss' Law. I've been thinking about different cases and conditions and I have been confused by the actual meaning of Electric flux being 0.

    1. The problem statement, all variables and given/known data
    If I consider a sheet of uniform charge per unit area (sigma) I can calculate the E due to the sheet at an arbitrary distance from it.
    I get E = sigma/2Epsilon0

    What if I now consider an area which is in the vicinity of the sheet but doesn't actually contain any charge?

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    ++++++++++++++++++++++++++++++++++++++++++

    What implications does this have on E?

    2. Relevant equations

    phi = closed surface integral of E.dA = sum of all charges inside/epsilon0


    3. The attempt at a solution

    Since the charge inside is 0, flux = 0 and the integral is 0.

    This makes sense since on the top side, E is 0 degrees to the surface normal, but on the bottom it's 180 degrees and cos 180 = -1. So the integral is expected to give me 0.

    What does this mean? Flux is 0 but clearly E isn't 0.

    If I place a test charge in the cylinder I'm considering it will move away from the plates. There IS E.
    So flux being 0 doesn't mean E is 0 too?

    I could draw parallels with the other Gauss' Law about magnetic flux. It's always 0 right? It's physical implication was that there is no magnetic monopole. Each B field line we think of must be either a loop or be extended to infinity.

    I tried to look at some general meaning of flux. I found that flux is sum of all inflow - sum of all outflow.
    So, this result when we look at a cylinder far away from the sheet is expected as the flux in and out are equal. This means ther are no field lines being "created" or "destroyed" i.e. there is no charge.

    But then when I started thinking about Faraday cages I got really confused.

    Does this mean that this law can never tell us the E ever. We can't calculate the electric field strength in a region using this law. Since, when we use this law, we don't consider other charges that could be nearby?
    Like in this case we didn't look at the fact that there was a charged sheet nearby.
    Is there any truth to this? Or am I doing it wrong? amidoinitrite guise?

    If I have, say, a point charge somewhere. And very far away I take a Gaussian sphere. The sum of all charges in is 0. So flux is 0. But the E isn't 0 since if I place a test charge there, it will experience a force.

    tl;dr
    Does this mean that this law can never tell us the E ever.
     
  2. jcsd
  3. Jul 20, 2010 #2
    Yes, the law doesn't tell us much about the E-field in the space. It only tells us about the flux. The Gauss law is a convenient way to calculate E in cases where symmetry is present.

    In your example, the chosen surface is out of the charged plane, and the flux through it is obviously 0. Now if we remove the charged plane, there will be no charge in the space, which means no electric field, and the flux, again, is 0. Therefore, all we know is the flux, not the E-field. The electric flux = 0 only means that there is no charges inside the surface (or the total charge = 0).

    To calculate E-field, we must know the distribution of the charges, or at least, some properties of E-field. Then we have two ways to calculate E-field: using the original formula of E, or exploiting the symmetry and using Gauss law.
     
    Last edited: Jul 20, 2010
  4. Jul 20, 2010 #3

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    Thanks bro. So I guess I was right after all. I don't have a professor or colleagues to go to when I can't solve a problem so you will be seeing a lot of me in the future.
     
  5. Jul 20, 2010 #4

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    I think hikaru1221 summed things up pretty well, but allow me to elaborate on a couple of key points at the risk of being redundant.

    Gauss' law is always valid in electrostatics. You can always use it to find the charge enclosed within a surface, if you happen to know what E is.

    That being said, it is only useful in determining E in a few special circumstances listed below. Note that all the circumstances below rely on symmetry.

    (a) Finding E inside or outside an object that has spherically symmetrical charge distribution. No other charges can be present, because that would disrupt the spherical symmetry (unless those other charges are also spherically symmetric and share the common center)! Spheres, spherically symmetrical shells and point charges typically fit into this category (assuming spherical charge symmetry and no other charges are present). Note that the charge distribution does not necessarily need to be uniform. It only needs to be spherically symmetric.

    (b) Finding E inside or outside a an infinitely long object with infinitely cylindrical charge distribution. No other charges can be present because that would disrupt the cylindrical symmetry (unless the other objects are coaxial and share the same symmetry). Infinitely long wires, infinitely long cylinders and infinitely long cylindrical shells typically fit into this category (assuming cylindrical charge symmetry and no other charges around). Note that the charge distribution need not be uniform -- it just needs to be infinite in length and cylindrically symmetric. Although technically the wire or cylinder needs to be infinitely long, Guass' law is often used as an approximation for long wires and long cylinders -- but realize it is just an approximation.

    (c) Finding E inside or outside of an infinitely long plane with unchanging charge distribution along the plane. In other words, the object can have some thickness, and the charge distribution need not be uniform as a function of depth. But the charge distribution as a function of depth must be the same (whatever that might be) at one point on the plane as any other point on the plane with the same depth. No other charges can be present unless they share the same planar symmetry. Although technically the plane must be infinite in size, Guass' law is often used as an approximation for large planes, where the distance from the plane is much much smaller than the length of the plane -- but realize it is just an approximation.

    And that's pretty much it.

    In your example, you have a charged sheet, but part of the sheet has zero charge and other parts do not. This violates the symmetry rule, so you won't be able to use Guass' law to find E. But that doesn't mean Guass' law is invalid. It is still valid! And you can still use it to find the enclosed charge, if you happen to know what the E field is.

    In your example, you will find some electric flux through some of the sides of the Gaussian pillbox. This flux would be zero if your example showed planar symmetry, but since your example does not have this symmetry, the flux is not zero on any given side. However, some sides will have a positive flux, and other sides will have a negative flux. After integrating [itex] \int \mathbf{E} \cdot \mathbs{dA} [/itex], of each side of the pillbox, you will find that some sides are not zero. But when you add the results all the sides together you will get zero (assuming the charge inside the pillbox is zero)! So Guass' law still works -- you just can't use it to find E here.
     
    Last edited: Jul 20, 2010
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