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Griffiths induced charge

  1. Mar 10, 2008 #1
    [SOLVED] Griffiths induced charge

    1. The problem statement, all variables and given/known data
    I am so insanely confused about induced charge.

    Griffiths does this problem in which a charge q is held a distance d above an infinite grounded conducting plane and then he calculates that the induced charge on the plane is -q (section 3.2.1). But if you start with just your charge q far away from the neutral conducting plate and then bring in the charge q to the distance d, you start out with total charge q and end up with total charge 0. How in the world is this not a violation of charge conservation?!? Please help me -this is a serious contradiction.

    Also, in problem 3.37, I was able to calculate the induced surface charge in part b but then part c really threw me off. Apparently I have to integrate over both spheres to get the total charge which is absolutely insane because before the person glued the charge on the outer sphere there was only that much charge "in the world" and then suddenly all this new charge appears on the inner sphere? WHAT IS WRONG HERE?!??!

    2. Relevant equations



    3. The attempt at a solution
     
    Last edited: Mar 10, 2008
  2. jcsd
  3. Mar 10, 2008 #2

    Dick

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    The conducting plane is still overall neutral. It simply concentrates opposite sign charges below the charge q in such a way as to produce the same effect outside of the plane as a image charge -q located symmetrically in the plane. Which it has to do to keep the E field normal to the plane at all points. The image charge isn't a real charge. Get over it. They are just saying you can replace the neutral plane with the image charge and get the same solution outside of the conductor. The field in the conductor is quite different than with the -q image charge. This is not a SERIOUS CONTRADICTION.
     
    Last edited: Mar 10, 2008
  4. Mar 11, 2008 #3
    That's what I thought before, but problem 3.37c threw me off. The total charge will then just be the charge on the outer sphere, right, since the inner sphere must be electrically neutral like Dick said, right?
     
  5. Mar 11, 2008 #4

    Dick

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    What is problem 3.37?? Can you quote it?
     
  6. Mar 11, 2008 #5
    Just find Q total in terms of V0 and then use that to find a ration between V and V0, you will see that it is not insane, but simply a limit case
     
  7. Mar 11, 2008 #6
    Problem 3.37
    A conducting sphere of radius a, at potential V_0, is surrounded by a thin concentric shell of radius b, over which someone has glued a surface charge [itex] \sigma(\theta) = k \cos \theta [/itex]. Parts a and b ask you to find the potential in each region and the induced surface charge [itex]\sigma(\theta)[/itex] on the conductor. Part c says find the total charge of the system.

    Parts a and b were easy. Part c is just confusing me. If what Dick said above is correct, then the inner sphere should be neutral and you should only have to integrate over the "glued" charge. But I found this solution http://ophelia.princeton.edu/~page/phy304/probsets/ps3_soln.pdf that says we need to integrate over both the inner and outer sphere, which IS a CONTRADICTION to what Dick said.
     
    Last edited: Mar 11, 2008
  8. Mar 11, 2008 #7
    Do consider that he said PLANE not sphere... there is a difference
     
  9. Mar 11, 2008 #8
    OK. But then can someone please explain to me why charge conservation holds in the case of Problem 3.37?
     
  10. Mar 11, 2008 #9
    Look at Example 3.2 and Exercise 3.8 they show the different options
     
  11. Mar 11, 2008 #10
    I looked at both of them and I still do not see the answer to question.
     
  12. Mar 11, 2008 #11
    That there are multiple possible positions for the image charges in a sphere. they are conserved even though they are positioned differently. You also should be be so strict on the conservation of charges when it comes to images charges as the actual position of charges is completely different to where we make them to be
     
  13. Mar 11, 2008 #12
    I don't see how that answers my question about Problem 3.37 which has nothing to do with image charges (unless by image charges you mean induced charges?)
     
    Last edited: Mar 11, 2008
  14. Mar 11, 2008 #13

    Ben Niehoff

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    Emphasis mine:

    The key word here is "grounded". What does it mean for an object to be "grounded"? It means that that object is held at zero potential, no matter what. And how would you accomplish that? By connecting the object (via a small wire of negligible size) to an infinite reservoir of charge, such that the charge from the reservoir can move into the object to precisely cancel any possible voltage changes. And when you have an infinite reservoir of charge in a problem, charge conservation is violated, yes. Griffith's is a great book, but this is one point on which Jackson actually explains something better: that is, exactly what it means to hold an object at a constant potential. It is a mathematical idealization that does not occur in reality.

    The induced surface charge on the infinite plane is quite real, and the total charge in the entire system (point charge + infinite plane) is zero. You also asked, however, what happens if I move the point charge infinitely far away? But that is a bit of a trick. The fact is, you can't move "infinitely far away" from an infinite conducting plane!

    For any finite distribution of objects, the further you are away, the smaller they look. More precisely, the solid angle subtended by the objects diminishes with distance (it goes as 1/r^2, in fact). As you get further and further away, any finite charge distribution will tend to look like a single point charge (if you read in Griffith's about the multipole expansion, this is essentially saying that very far away from a distribution of charge, only the monopole term is important).

    But for an infinite distribution of objects, the same logic does not hold. An infinite plane will always subtend 2*pi steradians, no matter how far away you move. This is, in fact, the very reason that the electric field of an infinite charged plane is constant; the plane does not diminish in apparent size as you get further from it.

    In other words, there is nowhere in the universe you can go where the effects of an infinite plane are not felt. Your point charge is always "sufficiently close" to the plane to induce an equal and opposite image charge on it, because the problem looks identical at all scales.


    For more fun, consider what happens as you get further away from an infinite cylinder/. It shrinks in one dimension, but not the other! So its apparent size goes as 1/r...
     
    Last edited: Mar 11, 2008
  15. Mar 11, 2008 #14

    Ben Niehoff

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    Careful here! Potential is not the same as charge!

    1. An object can be at a potential V != 0 and still have zero charge.

    2. An object can be at V = 0 and have net charge.

    3. An object can be at a constant potential and have a non-constant distribution of charge.

    4. An non-conducting object can have a non-constant potential and yet still have a uniform distribution of charge.

    Just because V has some value does not imply anything about Q.
     
  16. Mar 11, 2008 #15

    kdv

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    You are right and I deleted my post in case it could cause some confusion. But what I meant is the folowing: If you have a conducting sphere grounded at V_0 and you are taking the potential at infinity to be zero, then would you say that this is possible even if there is no charge on the sphere?
     
  17. Mar 11, 2008 #16

    Ben Niehoff

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    What does "grounded at V_0" mean? Either it is grounded, or it is held at V_0. "Grounded" quite literally means "held at zero potential".

    If the sphere is at potential V_0 and the potential goes to zero at infinity, then the sphere must have a net charge. You can find the surface charge density by solving Laplace's equation inside and outside the sphere, and taking the normal derivative of the potential at the surface of the sphere. More precisely,

    [tex]\sigma = \left -\epsilon_0 \; \vec n \cdot \vec \nabla\Phi_{out} \right|_S[/tex]

    where n is the unit outward normal and S is the surface of the conductor. Note that the potential is always constant inside a conductor, which is why only the outside potential appears in the equation.
     
  18. Mar 11, 2008 #17

    kdv

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    Sorry, I meant "the conducting sphere is at a potential V_0".
    This is what I meant in my initial post: that in that case (and assuming that as usual, V is chosen to be zero at infinity) then there has to be a net charge on the sphere. I think this is the whole point that caused confusion to the OP.
     
  19. Mar 11, 2008 #18
    I suspected Griffiths was not telling the whole story. The answer to my question about Problem 3.37 is this: gluing the charge to the outer sphere causes charge to flow into or out of the infinite charge reservoir in order to maintain the constant potential. Since the inner sphere IS a finite object, we could move the charge in from infinity and glue it to the outer sphere inducing surface charge on the inner sphere. If we DO NOT ACCOUNT FOR THE CHANGE OF CHARGE IN THE RESERVOIR THAT BEN SPOKE OF, then charge conservation is "violated."
     
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