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Gauss' law

  1. Apr 16, 2012 #1
    Is proven in my book using field lines. I stumbled upon something that I however can't understand. They say that a charge outside a closed surfrace contributes nothing to the total flux out of the surface since every field line that goes into the surface also comes out. But what if it terminates on a charge inside the closed surface? Like if you have a negative charge outside your closed surface and a positive charge inside it.
     
  2. jcsd
  3. Apr 16, 2012 #2
    Then the charge inside (be it negative or positive) is contributing to the flux through the surface. There is an enclosed charge so Gauss' law holds. The book is probably just trying to explain that any lines that enter the surface, but do not terminate inside it, must also exit and thus, contribute nothing towards the electric flux. This becomes an important concept when you begin to study Gauss' Law for Magnetism.
     
  4. Apr 17, 2012 #3
    hmm I don't think I understand. Doesn't Gauss' law say that all the lines coming out of a closed surface must come from the charge thus making you able to find the total charge inside the surface by looking at the total flux out of the surface?

    Well then when you have a charge outside, whose lines go through the surface and terminate on a charge inside, doesn't that give you a flux due to the charge outside the surface?
     
  5. Apr 17, 2012 #4
    No. The law of superposition states that the electric field of a material is the sum of the electric fields of all the individual particles contributing to the electric field. That means that for both particles you can look at the electric field independently of the electric field of the other particle. Gauss' law holds for both particles independently and the electric flux of a volume will only depend on the charge of the particle inside the volume.
     
  6. Apr 17, 2012 #5
    okay but then you must explain to me. Let's say we look at a +charge independently even though there is a -charge close to it. Then the flux through a closed surface is a measure of the amount of field lines passing through the closed surface, which by Gauss' law is only dependent on the charge inside the closed surface (right?).

    But the minus charge will have field lines going through the surface and terminating on the plus charge. Why do these not contribute to the flux? Evidently they are indeed going through the closed surface.

    Edit: Or is the E in the integral only supposed to be the E due to the charge we are confining?
     
    Last edited: Apr 17, 2012
  7. Apr 17, 2012 #6
    The net flux is indeed only dependent on the charge inside a closed surface.

    The net flux through a closed surface can be calculated for both particles independently (as of the law of superposition).

    Now I'm not entirely sure what situation you try to picture but when you place two particles next to each other, there will not only be field lines between them, but in all directions around the particles. At the side of the particle opposite to the particle the field lines will reach the other particles at an infinite distance.

    I found an image which might help illustrate it.
    image122.gif
    For any surface not enclosing either charge the integral [itex]\int_{A}\mathbf{E}\cdot\mathbf{n}dA[/itex] will be zero. This is easy to see for area's where [itex]\mathbf{E}[/itex] is uniform. Keep in mind that [itex]\mathbf{n}[/itex] is the outward unit normal vector.
     
  8. Apr 17, 2012 #7
    So say we have the combined field of the two charges above and we want to calculate the total charge inside a sphere that confines the negative charge. The E in the flux integral in Gauss' law, does that then refer to the combined field of the two charges or just the field that the negative charge makes?
     
  9. Apr 17, 2012 #8
    It's actually the other way around. You can't actually measure electric fields, there's no device that does that.
    Usually you know the charge that generates the electric field. Then you draw a closed surface around it so you can compute the electric field. The electric field of a charge is just pointed outwards (or inwards, depending on the sign of the charge) and is the same in all directions.
    2e.GIF

    Gauss' theorem, also known as the divergence theorem, says that the net flux of a vector field in a volume is equal to the divergence of that vector field integrated over that volume, or: [itex]\int_{A}\mathbf{E}\cdot\mathbf{n}dA=\int_{V}\nabla\cdot\mathbf{E}dV[/itex].

    The divergence of the electrical field is the charge density divided by the vacuum permittivity: [itex]\nabla\cdot\mathbf{E}=\dfrac{\rho}{\epsilon_{0}}[/itex].
    The charge density integrated over a volume is equal to the total charge enclosed within that volume: [itex]\int_{V}\nabla\cdot\mathbf{E}dV=\int_{V}\dfrac{ \rho}{\epsilon_{0}}dV=\dfrac{Q_{enc}}{\epsilon_{0}}[/itex].

    Now when we choose our volume in a convenient way such that [itex]\mathbf{E}[/itex] is always perpendicular to the surface of our volume, we will be able to easily obtain the electrical field.
    For a charged particle our best bet would be a sphere centered on the charge. We already solved the right hand side of the equation, so we only need to solve the left hand side. Since we know that both [itex]\mathbf{E}[/itex] and [itex]\mathbf{n}[/itex] point in the same direction and since we know that [itex]\mathbf{n}[/itex] only has an r-component, the integral becomes [itex]\int_{A}\mathbf{E}\cdot\mathbf{n}dA=\int_{A}E_{r}dA=E_{r}A=E_{r}4\pi r^{2}[/itex].

    Thus [itex]E_{r}4\pi r^{2}=\dfrac{Q_{enc}}{\epsilon_{0}}[/itex] or [itex]E_{r}=\dfrac{Q_{enc}}{4\pi r^{2}\epsilon_{0}}[/itex].
    Since the electric field only points in the r-direction, we can then obtain the electric field by multiplying with the outward unit normal vector: [itex]\mathbf{E}=E_{r}\mathbf{\hat{r}}=\dfrac{Q_{enc}}{4\pi r^{2}\epsilon_{0}}\mathbf{\hat{r}}[/itex]
     
    Last edited: Apr 17, 2012
  10. Apr 17, 2012 #9
    Okay, so as I stated, the field you calculate will be the field only due to the confined charge, not the total field of all the charges you are dealing with.

    So let's just for convenience get it perfectly straight. Say I have a positive and a negative charge. I want to calculate the field that the positive charge makes. Therefore I confine it within a sphere and use Gauss' theorem to calculate the field.

    Now the calculated field will not be the total field of the positive and negative charge, but only the field of the positive, i.e. a field with lines going out radially.

    But then.. What on earth is the point of saying that charges outside the sphere do not contribute to the field? Because of course they will not, since you are deliberately only looking at the field for the confined charge. Am I missing something here?
     
  11. Apr 17, 2012 #10
    You are missing the law of superposition. In your example you are only calculating the electrical field due to one particle, but for the total electrical field you also need to calculate the electrical field due to the other particle. The law of superposition says that the total electric field is the sum of the electrical fields of the individual particles. When you add these fields and take into account that they both have a different position, you'll see that field gets distorted.

    Contributing to the field is not the same as contributing to the flux.
     
  12. Apr 17, 2012 #11
    Well I meant contributing to the flux. I don't see the point of saying that particles outside do not contribute to the flux. Because aren't you looking at the fields of your charges confined, when you do the flux integral?

    Let me do it again:
    You have a plus and minus charge. Let us say I want to calculate the combined field. I can either confine both of them within a sphere and then then do the flux integral to calculate their total field. Or I can confine them in a sphere independently and then add the two fields together in the end. But in that proces I am pretending that the field of the other charge outside the sphere is nonexistent. So whats the point of saying that charges do not contribute to the flux, because of course they wont, since you are pretending they are not there when you do the integral...?
     
  13. Apr 17, 2012 #12
    You aren't "pretending" that the other charge isn't there when you use a Gaussian sphere that only encloses one charge, you're simply acknowledge that it doesn't contribute to the flux through that Gaussian surface. So the fact that a correct statement of Gauss's law writes that the electric flux is equal to the charge enclosed, you can consider a definition of Gauss's law. To prove that Gauss's law is true even when there are other charges outside your Gaussian sphere, you can do proofs with the field lines, etc. But once you know it to be true, you only use the charge enclosed to calculate the flux.
     
  14. Apr 17, 2012 #13
    So let's say again you have a plus and a minus charge, the plus charge confined within a sphere. What you are saying means that:

    Calculating the flux integral will be the same using a field composed of a superposition of the plus and minus particle's field and a field only generated by the plus particle.

    But that is just weird. The field's look completely different. And whilst it's true, that some field lines that enter from the minus particle do go out again, what then about the ones that terminate on the plus charge?

    I think we are once again back to the first post of this thread :(
     
  15. Apr 17, 2012 #14
    That is exactly correct. Your disbelief comes from the fact that based on your visualization, there's no reason for this to be true. And in a sense, you're right -- there's no reason you would be able to intuitively figure this out based on the shape of the field lines from the resultant field. If you don't believe it, I encourage you to work it out: write the electric field at a given location on the Gaussian surface as a superposition of the two Coulomb fields, and take the surface integral. It's not trivial, but can be done. If that's too much work, well, just take it on faith that it's true, if the fuzzy "field line" proof of Gauss's law doesn't do it for you.

    What happens to the field lines outside of the Gaussian sphere is irrelevant, since all that matters is the flux through the surface.
     
  16. Apr 17, 2012 #15
    Okay, so we've come so way. It's just still weird for me that charges outside contribute nothing whatsoever.

    1) Like, as far as I can see the symmetry that Gauss' law states is the fact that even though the strength of our field decreases proportional to 1/r^2 then the area of any surface confining a charge also increases proportional to r^2 thus making the two things cancel ellegantly.

    But when you have a charge outside it totally changes the field at the point of the surface, so what is it that makes it still hold? Like what is it that makes this changing of the field at the surface point contribute nothing to the total flux?

    Maybe what I say in 1) is wrong and I must invoke another way of thinking?
     
  17. Apr 17, 2012 #16
    Gauss's law is not supposed to be a throwaway statement. It is hugely powerful. If you don't find it weird and amazing when you first learn about it, you don't really understand it yet.

    This is a true statement about Gauss' law, but it's not really the definition. There is really no way to prove Gauss' law from any sort of first principles: it is the first principle. It's a fundamental fact about nature. Given that interpretation,

    The property that makes it this way is just the definition of the surface integral. If you aren't happy with the conceptual statement that a field line that enters must also leave, can you be satisfied with the mathematical one that if you draw a sphere that encloses no charge, and put a positive charge next to that sphere, the total flux through that surface from the positive charge is zero turns out to be zero, after doing the math?
     
  18. Apr 17, 2012 #17
    Okay, thinking it all through made a lot of things make sense, and I think I understand it now, at least understand it as much as I like.
    When trying to sleep one thing did however occur weird to me: Namely the way you define the flux as the dot product of the field and the surface elements at every point. Why is it that you have to invoke this dot product for Gauss' law to make sense? Like the total amount of field lines going out of surface will always be the same regardless of how they are oriented compared to the normals of our surface.

    Say you have a field due to some arrangement of charges, which is confined in a sphere. The field lines I imagine now look such that they don't go out radially but instead turn such that they are almost perpendicular to the radial normals of the sphere. Then the flux integral will be almost zero, and thus the charge inside must be the same?
     
  19. Apr 18, 2012 #18
    This is true and is what happens in electrodynamics. In that case the divergence of the field is zero. Stokes theorem states [itex]\oint_{\Gamma} \mathbf{E} \cdot d \mathbf{s} = \int_{A} \left(\nabla \times \mathbf{E}\right) \cdot \mathbf{n}dA[/itex] where [itex]\Gamma[/itex] is a loop enclosing an open surface A and [itex]\mathbf{s}[/itex] is the unit vector tangential to the surface. Now the Maxwell-Faraday equation says [itex]\nabla\times\mathbf{E}=-\dfrac{\partial\mathbf{B}}{\partial t}[/itex] and there are similar tricks to solve the equation.

    However not all kinds of electric fields are possible. What kind of field you get depends on how your problem is specified. In electrodynamics the electrical fields are going round as in the image below and in electrostatics the field will always radially point outwards from a particle (a charged sheet is made up of charged particles). Of course a combination is possible, for example in an electrical wire, but when you're talking electrostatics you can assume that [itex]\nabla\times\mathbf{E}=0[/itex] and you will not see the circular pattern.

    derham2.png

    But before you'll learn about electrodynamics, you'll probably first learn about magnetostatics. In magnetostatics, the magnetic field is dependent on the current density ([itex]\nabla\times\mathbf{B}=\dfrac{\mathbf{j}}{\epsilon_{0}c^{2}}[/itex]) and you use similar techniques to find the magnetic field.

    You don't need to worry about electrodynamics for now, just assume that [itex]\nabla\times\mathbf{E}=0[/itex] in case of electrostatics. After you have a deeper understanding of electrostatics you'll learn about magnetostatics and after that about electrodynamics.
     
    Last edited: Apr 18, 2012
  20. Apr 18, 2012 #19
    This is incorrect. If a field line is parallel to a surface, can you really say that it is "going out of" that surface? Of course not. The very definition of flux relates to only the field lines that pass through the surface. But if you're asking why it is the flux in particular that plays the part in Gauss' law, I'm not sure I have a better answer than that's just the way nature works.

    Again, think back to Gauss' law. The value of the flux integral does not depend on the arrangement of charges inside the sphere; it only depends on the net charge in your sphere. So, for example, if all of your charges were positive charges, then you amplify the total flux by counting the field lines from every particle, and none of those are diluted by field lines which point in the opposite direction.
     
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