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Griffith's Second Uniqueness Theorem

  1. Feb 10, 2008 #1
    [SOLVED] Griffith's Second Uniqueness Theorem

    1. The problem statement, all variables and given/known data
    I am having trouble understanding the Second uniqueness theorem in Griffith's Electrodynamics book which states that

    "In a volume V surrounded by conductors and containing a specified charge density rho, the electric field is uniquely determined if the total charge on each conductor is given. (The region as a whole can be bounded by another conductor, or else unbounded.)"


    2. Relevant equations



    3. The attempt at a solution
    As stated, that theorem makes absolutely no sense to me, especially the part in parenthesis. I really don't even know how to ask a question about it because it simply makes no sense. Can someone give me a link to a site where it is explained better. If anyone has the book, look at Figure 3.6. What does the darker and the lighter shading represent? Is the "integration surface" also a conductor?
     
    Last edited: Feb 10, 2008
  2. jcsd
  3. Feb 10, 2008 #2

    quasar987

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    Here's how I understood this at the time.

    The volume V refers to a region of space where there is no conductor.

    So pick and insulator of your choice.

    Start by filling the universe with this insulator of charge density rho. Then start carving holes in the insulator like holes in a piece of swiss cheese and put some arbitrarily charged conductors in there. This is the situation he refers to at the end by saying the region in unbounded.

    The situation in which the region in bounded corresponds to taking another step in our construction where we now carve out allll but a bounded region of insulator/cheese, and replace it with a conductor.
     
    Last edited: Feb 10, 2008
  4. Feb 10, 2008 #3
    When you carve out the conductors, are you carving them inside the volume V or outside?
     
  5. Feb 10, 2008 #4

    quasar987

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    I don't understand your question, but there seems to be a misunderstanding.

    You are a god, and you start by filling a void universe with insulator.

    This is why I don't understand your question.. there is no inside or outside.. just insulator everywhere!

    Then you may decide to place a charged metallic ball of radius R somewhere.. so you magically make a ball of radius R of insulator disappear where you want the metallic ball and you teleport it in the hole.

    etc.
     
  6. Feb 10, 2008 #5
    You said the volume V refers to a region of space where there is no conductor. But I guess you meant there is no conductor there until you put it insert one.

    I think the theorem would make much, much more sense if I changed "surrounded by" to "surrounding". That also seems to be the situation in your example. Is that a mistake in the theorem?
     
  7. Feb 10, 2008 #6

    kdv

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    Sounds like you are right. It only makes sense to me if the volume is surrounding the conductors. and fig 3.6 then makes sense.
     
  8. Feb 10, 2008 #7
    On second thought, I think "containing" would be a better word. Volumes don't really surround. But in any case, it must be wrong as stated.
     
  9. Feb 10, 2008 #8

    kdv

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    Yes, you are right, again.
     
  10. Feb 10, 2008 #9
    Hmmm. Sorry to keep changing my mind, but I am starting to think "surrounded by" is actually right. If you look at Figure 3.6, notice that the S has two arrows, one pointing to the outer boundary and one pointing to the conductor containing Q_2. This indicates that, in Figure 3.6, the conductors contains Q_1,Q_2,Q_3, and Q_4 ARE NOT PART OF THE VOLUME V. Now, regardless of whether the outer boundary is a conductor or at infinity, the volume will indeed be surrounded by conductors...the reason I was confused is that it is surrounded by them both on its inside as well as its outside (if the boundary is not infinity). "surrounded by" conductors just means that all boundaries of the volume are also boundaries of conductors.
     
  11. Feb 10, 2008 #10

    quasar987

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    After you're done sowing your conductors completely and you're satisfied with your creation, call V the region filled with insulators.
     
  12. Feb 10, 2008 #11

    quasar987

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    You've got it. Or at least, now your interpretation and mine seem to concur.
     
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