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Conducting shells

  1. Nov 2, 2008 #1
    A spherical conducting shell has an inner radius R1 and outer radius R2

    There is a charge Q1 spread uniformly on the inner radius

    THere is a charge Q2 spread uniformly on the outer radius

    what is the potential for a distance R1<r<R2?

    What is the potential for a distance r<R1?

    Since the potential is continuous between surfaces, the potential the outer surface and inner surface must be equal at the boundary

    I'm not sure where to begin with this question. I was thinking about treating the surfaces as point charges but we cant do that in this case since we want to know the potential between the shells. Any advice?
     
  2. jcsd
  3. Nov 3, 2008 #2

    gabbagabbahey

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    My advice is to exploit the spherical symmetry in the problem, by using Gauss' Law to calculate the electric field and then integrating it to find the potential.
     
  4. Nov 4, 2008 #3
    i forgot to mention that the potential on R2 is grounded

    the potential inside r<R1 should be [tex]V=\frac{KQ_1}{R_1}[/tex] since the potential inside should equal the potential at the surface.

    The potential at R2 is zero, but must increase to [tex]V=\frac{KQ_1}{R_1}[/tex] at R2 right?
     
  5. Nov 4, 2008 #4

    gabbagabbahey

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    Assuming you meant to say that V goes from zero at R2 to [tex]
    V=\frac{KQ_1}{R_1}
    [/tex] at R1, then yes...but to determine the functional relationship of how V increases in this region, you will need to use gauss' law
     
  6. Nov 4, 2008 #5
    i'm not sure how to take into account that V=0 at R2

    [tex]V=-\int_{R_1}^r \frac{kQ_1}{r^2} = kQ_1 \left( \frac{1}{r} - \frac{1}{R_1} \right) [/tex]

    ?
     
  7. Nov 4, 2008 #6

    gabbagabbahey

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    What you've computed is the potential due to just the inner shell of charge, not the potential due to the total charge configuration...you can find that by using the superposition principle...what is the potential at r due to the outer shell of charge?
     
  8. Nov 4, 2008 #7
    potential due to the other shell is zero isnt it?

    if we just looked at the outer shell, E=0 inside the shell and since the potential at the outer shell is zero, then the potential inside the shell should be zero as well.
     
  9. Nov 4, 2008 #8
    don't forget induced charges....
     
  10. Nov 4, 2008 #9

    gabbagabbahey

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    E is zero inside the shell, but the potential isn't: Remember, [itex]V_{total}(R_2)=0[/itex] while [itex]V_{duetoQ_1}(R_2)=kQ_1 \left( \frac{1}{R_2} - \frac{1}{R_1} \right)[/itex] and so [itex]V_{duetoQ_2}(R_2)=V_{total}(R_2)-V_{duetoQ_1}(R_2) \neq 0[/itex]. [itex]V_{duetoQ_2}(r)[/itex] will be constant, since E=0 inside the outer shell, but it will not be zero.
     
  11. Nov 4, 2008 #10

    gabbagabbahey

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    There is no need to consider induced charges here, since the charge density on all surfaces in the problem is specified.
     
  12. Nov 4, 2008 #11
    so the potential is be:

    [tex]
    V(R_2)=-kQ_1 \left( \frac{1}{R_2} - \frac{1}{R_1} \right)
    [/tex]


    [tex]
    V= kQ_1 \left( \frac{1}{r} - \frac{1}{R_1} -\frac{1}{R_2} + \frac{1}{R_1} \right)
    [/tex]

    [tex]
    V=kQ_1 \left( \frac{1}{r} - \frac{1}{R_2} \right)
    [/tex]

    the potential doesnt depend on Q2?

    it's also curious that there IS potential on the outer shell we it is grounded (to cancel out the potential from the inner shell)
     
  13. Nov 4, 2008 #12

    gabbagabbahey

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    Your result is correct....V(r) does not depend on [itex]Q_2[/itex] in the region [itex]R_1<r<R_2[/itex]....outside the outer shell however it would depend on Q2.
     
  14. Nov 4, 2008 #13
    so the total potential at the outer shell is zero, but in order for this to happen, there must be a potential on the outer shell to cancel the inner shell?

    is Q1 and Q2 related in any way?

    should the two be the same?

    Just Q1=-Q2 (inside)

    so the E field seen from the outside is just kQ1/r^2?
     
  15. Nov 4, 2008 #14

    gabbagabbahey

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    Q1 and Q2 do not need to be related in any way.

    The total potential at R2 must be zero, and given that the potential at R2 due to the inner shell is non-zero, the potential due to the outer shell must also be nonzero at R2 in order for the two to cancel each other out.

    The Efield due to the inner shell is kQ1/r^2 for r>R2 but the total E-field at r>R2 is k(Q1+Q2)/r^2...(or were you asking about the region R1<r<R2?)
     
  16. Nov 4, 2008 #15
    i'm not sure i follow, here's what i'm thinking

    the inner shell has a charge of Q1 on the outside

    so similar to image charges, wouldnt the grounded shell have a charge -Q1 on the inside (and +Q1 on the outside) so when seen from r>R2, it's as if there was just a point charge of Q1 at the origin since the outer shell is grounded and the inner shell is charged.
     
  17. Nov 4, 2008 #16

    gabbagabbahey

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    Hmmm.. I seem to have misread the original problem:

    There is only a single shell in this problem. It is a conductor, so the field in the region R1<r<R2 must be zero and the potential must be constant in this region! Moreover, Q1 must be zero (as a quick application of Gauss' law shows that E=kQ1/r^2 but E=0)!

    Also the field inside r<R1 is zero, so the potential in that region is constant too!

    The potential must be continuous everywhere, if the outer surface is grounded, then V(R2)=0, and hence V(r)=0 for [itex]r \leq R2[/itex]

    The field outside is due entirely to Q2 since Q1 is zero.
     
  18. Nov 4, 2008 #17
    oh, i'm still confused about when there are two shells (the problem we were answering before)

    was what i wrote above true? Because i dont see how the total E-field at r>R2 is k(Q1+Q2)/r^2
     
  19. Nov 4, 2008 #18

    gabbagabbahey

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    If the shells are assumed to be very thin, then they are treated as each having a single surface rather than an inner and an outer surface for each shell...that is what I had originally thought the problem was.

    If you want two shells of finite thickness, you need to say whether Q1 is on the inner or outer surface of the shell, and similarly for the outer shell and Q2. You also need to specify the thickness of each shell.

    edit- or are you thinking of the case where Q1 is the net charge of the inner shell and similarly for the outer shell?
     
  20. Nov 4, 2008 #19
    right, for the problem with two very thin shells, the result that was derived was correct:

    V=kQ1/R1 for r<R1

    and

    V=kQ1(1/r - 1/R2) for R2>r>R1

    but i'm saying that E for r>R2 is kQ1/r^2 and not k(Q1+Q2)/r^2

    going by the reasoning that:

    is this valid?
     
  21. Nov 4, 2008 #20

    gabbagabbahey

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    No, the outer shell has charge Q2.

    There is no "inner" or "outer" surface of the R2 shell, since it is assumed to be "very thin".


    If you instead looked at a shell of finite thickness, then yes a charge of -Q1 would have to be induced on its inner surface, but(!) if its net charge is given as Q2, then the charge on its outer surface would have to be Q1+Q2 not just Q1 (unless Q2=0 i.e. for a neutral outer shell)
     
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