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Conceptual questions on electrodynamics

  1. May 20, 2008 #1
    Hi guys

    I have some questions about electrostatics, and I hope you can help me. Here they are:

    1) Please take a look at the following example: http://books.google.dk/books?id=YkG...bXLVw66&sig=YiULcbQhTa7htSGbHJ4-D4ZBaJU&hl=da

    Here we are calculating the electric field E of a uniformly, polarized solid sphere. My question is: When it is uniformly polarized, there are no volume bound charges. Then how can we have an electric field inside the sphere? Isn't the setup equivalent of a spherical shell, which has E=0?

    2) I have two conducting metal plates with a distance d between them. I put a potential of 500 V on the plates, and I want to find the electric field. I use the following

    [tex]
    V = - \int_C^r {\overline E } \cdot d\overline l
    [/tex]
    where C is a reference point, which I set to zero, which is at one of the plates.

    I get that 500 V = -E*d. Why do I get a minus, when it should be positive (I know that it must be a positive number).

    3) Is it possible to have negative capacitance? I would personally think no, when thinking of the physical meaning of capacitance, but I need to be sure.
     
  2. jcsd
  3. May 20, 2008 #2

    Redbelly98

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    When I click that link, I get a page showing an entire book.

    Those +/- signs can be tricky. Just to be clear, you have chosen as reference point the plate at a lower potential, so the other plate is at +500V.

    The integral contains a dot-product of two vectors. If they are at 180 degrees to one another, the dot-product is negative and gives an overall positive result. Think about what those 2 vectors are, and more importantly what their direction is in terms of how we have set up the problem. I.e., does E point from + to - or from - to + in a capacitor?

    From the basic definition of capacitance,

    C = Q/V,

    A negative C implies that the electric field points the "wrong way" w.r.t. the + and - charges on a capacitor.

    Hope this helps, good luck.
     
  4. May 20, 2008 #3
    1) Please follow this link, page 6, example 2: http://www.phys.unsw.edu.au/~gary/2050Lectures_4.pdf

    2) When I am not told the direction of the E-field, then what am I to assume? I can't find the direction of E from V.

    And where does one usually place the reference points? Do we always go from the plate with lowest potential to higher or what?
     
  5. May 20, 2008 #4

    Doc Al

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    The field points from higher to lower potential. You (arbitrarily) defined your positive dl to point towards higher potential, so E will be negative.

    If you don't know which plate is at the higher potential, just give the magnitude of the field.
     
  6. May 20, 2008 #5

    Redbelly98

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    Okay, I see. You're original question is, why is there an E-field inside a uniformly polarized sphere, when there are no bound charges within its volume?

    Another contribution to E comes from the bound surface charge density. That is what gives rise to the E-field here. See Eqn. 4.11 and the discussion leading up to it.

    But you can. So the problem now is to find E, with V=500V as a given, and use the fact that E is uniform between the capacitor plates. (If that's not the problem you're asking, let me know.)

    Use the integral you gave before:

    [tex]
    V = - \int_C^r {\overline E } \cdot d\overline l
    [/tex]

    Since the integral is evaluated from the lower-potential electrode to the higher-potential electrode, [tex]d\overline l[/tex] points towards the higher electrode, which we know is at +500V relative to the lower (reference) electrode.

    In order to calculate the voltage to be our known positive value, E must point in the opposite direction as dl, namely from + towards the - (reference) electrode.

    References are often, but not always, placed at the lowest potential in a DC electrical circuit.
     
  7. May 20, 2008 #6
    1)
    That is exactly my question. The bound surface charges are on the surface of the sphere, i.e. we have no enclosed charges, which from Gauss' law gives us that E=0. This is why I cannot see that there should be an electric field inside the sphere.

    2) Ok, so E always points from higher potential to lower potential. This is good.

    Does the dl mean that we integrate in the same direction as E? So in this case from the higher-potential plate to the lower-potential plate:

    [tex]

    V = - \int_d^0 {\overline E } \cdot d\overline l = Ed ?

    [/tex]

    Thanks alot for helping me. This is really useful to me.
     
  8. May 20, 2008 #7

    Redbelly98

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    (1) From what I remember, no volume charge implies that [tex]\nabla \cdot E [/tex] is zero. E could still be nonzero.

    Otherwise how could you have an E-field between two capacitor plates, or at some distance from a point charge?
     
  9. May 22, 2008 #8
    The l means you can go anywhere, the dot product of [E and l] is what tells you if stuff is going up or down. If they point in opposite directions then the dot product of [E and l] will be the negative of [E and l] alone. Then you throw in the negative that was hanging out front to find the actual potential.

    For me it is easiest to always integrate in the direction of the electric field, then take the negative of that if the problem was actually asking about something going the opposite way. This works better for me rather than always having to double negative [- * -] everything that ends up being positive. However it does work out mathematically everytime if you stick to the form you gave, with the negative of E.




    Also, Gauss' Law does not state that the E is zero but that the Electric flux is zero. You do not have to have charges within an arbitrarily shaped closed surface to have an electric field, just to have net electric flux. The idea with Gauss' law is to use it to find out things when the situation is symmetric, there is not symmetry here because the sphere is polarized.
     
    Last edited: May 22, 2008
  10. May 25, 2008 #9
    I can't bring up the text. So I wonder if something may be getting lost in translation. But in English, "A uniformly charged sphere" is unfortunately, ambiguous as to whether the volume or the surface of the object is referred to.
     
    Last edited: May 25, 2008
  11. May 26, 2008 #10

    Doc Al

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    Maybe so, but this thread is discussing a uniformly polarized sphere, not a uniformly charged one.
     
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