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Dumb thought experiment

  1. Jun 28, 2007 #1
    suppose you have a sphere of nonconducting material with a point charge not in the center but in one octant and around it you have a conductor. if examined outside the conductor,will there be a stronger field near the octant where the charge is in the insulator?

    gauss' law says no, vivid imagination says yes
     
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  3. Jun 28, 2007 #2

    olgranpappy

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    outside the conductor in this case means inside the nonconducting sphere and Gauss' law does not say "no." You are confused.
     
  4. Jun 28, 2007 #3
    no you are confused for 0<R<5 is the insulator for 5<R<10 is the conductor. the question concerns for R>10
     
  5. Jun 28, 2007 #4
    Agreed, Gauss' Law does not say no, although I can see where you are getting this idea. Gauss' Law says the TOTAL electric field through a closed surface is the same, no matter what the charge distribution (assuming the total charge enclosed is the same in all of the cases). However, this does not mean that the electric field through each unit area element is the same. You can only say this in the case of a sphere concentric with a spherical charge distribution because of the symmetry. Hope I've help!
     
  6. Jun 28, 2007 #5
    but then how does a faraday cage work, you guys are forgetting that there can be no field through a gaussian surface inside a conductor
     
  7. Jun 28, 2007 #6
    Ah, sorry - my bad, I mis-interpretted the situation. :redface: OK, at this point, I'm going to say I don't know what happens or what the answer is; but I'm going to try it. Hopefully, I've interpretted your question correctly this time!

    So, the charge creates an electric field, which isn't symmetric with the conductor which surrounds it. Due to the electric field, charge (of opposite 'sign' to that in the insulating sphere [- for simplicity, I'll assume the charge on the insulator is positive, so the induced charge on the inner surface of the conductor is negative]) is induced on the inner surface of the conductor. But, there is obviously more charge on the surface which is nearer to the sphere of charge than the surface further away, as the electric field is stronger. (See diagram for a very bad illustration!). Hence, as the conductor had no charge to start with, a positive charge must be induced on the outer surface, as charge can only exist on the surface of conductor. Also, as you correctly said, there is no E-field inside the conductor (net charge enclosed = 0). Hence, the charge on the outer surface can distribute itself evenly on the outer surface, as there is no E-field keeping it in place like on the inner surface. So I guess that says that the field isn't stronger near the octant where the charge is located. Which I guess is not what you'd initially think. As I said, I'm in no way an expert, and this is just my attempt to bundle my way through the problem (and most probably incorrect!), so it'd probably be a good idea to wait until a big-wig comes along - but I guess the fun of Physics is trying to work though a problem you don't really know!
     

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  8. Jun 28, 2007 #7

    Claude Bile

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    I think the result makes sense - charge on the conducting sphere will naturally even itself out. Think of the converse situation, charge clumping in one space in a conductor. This will create a potential inside the sphere, which will then cause current to flow until the voltage inside the conductor becomes constant (this is why you cannot get an E-field inside a conductor in a steady-state situation), at which point the charge distribution will be constant.

    Claude.
     
  9. Jun 28, 2007 #8

    olgranpappy

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    mind reader

    All you said in your original post was that you have a non-conducting sphere surrounded by a conductor. You didn't specify that the conductor was finite. And more importantly you did not specify that the outer surface of the conductor was spherical. I can't read your mind. In the case of a spherical conducting shell, yes the field outside the conductor for "R>10" is spherically symmetric and the potential is the same regardless of what octant you are in. So what? You can not make this claim if the outer surface of the conductor is not a spherical surface! And you did not mention this point in your question so you should not have expected anyone to assume this.
     
  10. Jun 28, 2007 #9
    this is the problem i was/am having. a single point charge has a field represented by a divergent vector field, you know arrows pointing away radially from it in all directions or field lines pointing away from it radially.

    two charges near each other do not cancel each other's field completely everywhere. on the axis of the dipole there is a field line emanating from the positive charge outward away from the negative charge, and into the negative charge there is a field line pointing towards the positive charge like this

    http://spiff.rit.edu/classes/phys273/manual/equipot2.gif

    if i put a gaussian surface around those two charges yes there will be zero net flux but tell me what the field will be like along the axis outside the surface? what does gauss' law say about the field at a single point on the axis? nothing and inspection say there is field there. so formally, how is that there is no field inside the conductor?

    edit

    i understand the physical principle, but obviously the vector field and math has to agree with the physical principal so how
     
    Last edited: Jun 29, 2007
  11. Jun 29, 2007 #10

    olgranpappy

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    more and more conductors

    Inside the conductor now, are we? With respect to your regions, do you mean inside the shell ("R<5") or literally inside the conductor ("5<R<10"). For R<5 there *is* a field. For 5<R<10 there is no field because the entire thing is at the same potential and all charge is on the surface thus the potential inside is everywhere constant so the field is zero (this regardless of the shape of the conductor, of course). Of course, you may find this reasoning circular, so let me give you the physical reason (I will just quote Landau vol. 8):

    "First of all, it follows from the fundemental property of conductors that, in the electrostatic case, the electric field inside a conductor must be zero. For a field [tex]\vec E[/tex] which was not zero would cause a current; the propagation of a current in a conductor involves a dissipation of energy, and hence cannot occur in a stationary state..."
     
  12. Jun 29, 2007 #11
    between 0<R<5

    first that wasn't a formal treatment at all, when i said formal i meant show me that the vector fields cancel with math.

    secondly did you not read my post? right at the bottom i mention that i know the physical reasons why it happens.

    edit

    im sorry 5<R<10
     
    Last edited: Jun 29, 2007
  13. Jun 29, 2007 #12

    olgranpappy

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    You obviously don't; for 0<R<5, as I stated in *my* previous post, the vector fields do not "cancel with math." There is an electric field.
     
  14. Jun 29, 2007 #13
    forgive me, i made the same mistake you did in your first post in this thread , i meant 5<R<10 and here the vector fields are supposed to cancel with math. the question has always been about fields inside the conductor.
     
  15. Jun 29, 2007 #14

    olgranpappy

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    okay. so then you don't want to start from the assumption that E=0 inside, you want to treat the problem in some other way. but then what properties do you propose we give to the region 5<R<10? We must treat it differently than the other two regions in some way as our theoretical starting point. what would you consider as an acceptable *starting point* for the proof that the electric fields cancel inside the conductor?
     
  16. Jun 29, 2007 #15
    obviously you can't start with the assumption that E=0 inside the conductor because that's what you're trying to prove. start with the same assumption i've made namely [tex]\vec{E}=k \frac{q}{R^{2}}\hat{r}[/tex] for a point charge.
     
  17. Jun 29, 2007 #16

    olgranpappy

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    this does not specify the problem completely. E.g., there will be induced charges on the surface of the conductor--Indeed (as we will see) induced in exactly the right way as to cancel the field inside. So, what I'm saying is: you need to specify the boundry conditions before the problem has a solution. As it stands you have only specified the boundry conditions on one boundry. How do you propose to treat the boundry that is the inner surface of the conductor?
     
  18. Jun 29, 2007 #17
    what do you mean by boundary conditions, do you intend to do an integral or something?
     
  19. Jun 29, 2007 #18

    olgranpappy

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    ...I intend to solve a differential equation and thus we must specify boundry conditions... or rather, since you are posing the problem, you must specify boundary conditions... or should I just take a guess?
     
  20. Jun 29, 2007 #19
    show me the de
     
  21. Jun 30, 2007 #20

    olgranpappy

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    [tex]-\nabla^2\phi=4\pi\rho[/tex]
     
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