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Concentric Metal Sphere - Gauss's Law

  1. Dec 14, 2011 #1
    1. Please see the attached problem, essentially a metal ball with charge Q1 is surrounded by a grounded shell q which in turn is surrounded by a further shell with charge Q2

    Q1. The first question is what is the charge induced on q by Q1 and Q2
    Q2. The second question is how is the charge distributed between the inner and outer surfaces of q
    [/b]


    2. None



    3. My problem is I am not sure what the effect of grounding is. From the information given I have assumed the following:

    qinner = Q2-Q1
    qouter = Q1-Q2
    qnet = qinner+qouter=Q2-Q1+Q1-Q2=0

    This I believe would apply to an ungrounded shell so what is the effect of grounding if the net effect = 0?

    200v2vc.jpg
     
  2. jcsd
  3. Dec 14, 2011 #2

    Simon Bridge

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    What is the field inside a charged sphere?
    From that, what is the contribution due to the outer sphere?

    Thus, what is the electric field on the the center sphere like?
    How would the charges in the middle sphere arrange themselves to that field?
     
  4. Dec 14, 2011 #3
    Eliminate the outer spherical shell. Remove the ground. Place a charge Q on the inner shell's outer surface then a charge -Q will form on the inner surface of the outer shell and a charge Q will be induced on the outer surface of the outer shell? Now ground the outer shell and all that changes is the positive charge Q will be drained off the outer shell while the charge -Q remains on the inner surface of the outer shell and the charge Q remains on the outer surface of the inner shell? Now add the outer shell and nothing changes?

    Hope this helps.
     
  5. Dec 15, 2011 #4
    My understanding is this,

    Inside a charged sphere the field E=Q/ε04∏r2 i.e. the net flux through any closed surface is equal to the net enclosed charge - Gauss's Law - rearranged to solve for E
    The contribution due to the outer sphere is essentially the same but the outer and inner polarities are reversed (because the outer sphere is outside?)

    The electric field on the centre sphere is 0 as it is assumed to be in equilibrium
    The -ve charges in the middle sphere are attracted to both the positive of the inner sphere and positive of the outer sphere (see my original solution).

    I'm not sure where you're going with the first three questions but essentially the last question I have attempted to answer in my original working. Essentially what I don't understand is that the ground can supply charge to balance out the effect of the two charged objects surrounding the inner sphere so the -Q's on each side are sunk to earth while the +Q's are provided with electrons, thus the net charge distribution = 0 OR are only the +Q's provided with electrons and the -Q's remain on the surfaces leaving -Q1 on the outer sphere and -Q2 on the inner sphere.

    Hope you can see where I am confused, thanks for your input

    If the outer shell is charged then the same effect will happen will it not? so this time a -Q will form on the outer surface (shown in original working) and Q will form on the inner surface i.e. it is the other way around because this time the charge is outside the sphere.

    Thanks
     
  6. Dec 15, 2011 #5
    Just a sec... Because the inner metal ball induces a charge +Q on the outer surface of the middle sphere and this is negated due to grounding, are you saying there is no electric field between the middle sphere and outer sphere thus no potential difference is created?

    i.e. qinner = -Q1
    qouter = 0

    Thanks again
    Dan
     
    Last edited: Dec 15, 2011
  7. Dec 15, 2011 #6

    SammyS

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    I don't know if he's saying that or not, but yes, it's true that there is zero electric field between the grounded sphere and the outer sphere.
     
  8. Dec 15, 2011 #7

    Simon Bridge

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    The electric field inside a spherical shell of uniform charge distribution is zero. By Gauss' Law: there is zero charge inside the shell.
    I hadn't got that far yet :)

    Q1(part): what is the charge induced on q by Q2?

    By the superposition principle, you can take each field (of the two charged sphere's) separately: therefore, the only field you need to account for is that of the inner sphere.

    How would this affect the charges in the middle sphere, if it were not grounded?
    eg. Faraday's "ice bucket" experiment.

    Then consider what effect grounding has on the situation.

    [edit]
    re-reading your answer sheet from post #1... the effect of grounding is where you are confused.
    Consider a related experiment:

    Charged object A is brought close to one side of neutral conductor B, glued to an insulated surface so it cannot move.
    The opposite side of B is briefly grounded, then A is removed to a far location.
    The charge on B is measured - what do we find and why?
     
    Last edited: Dec 15, 2011
  9. Dec 16, 2011 #8
    The charge on B would be equal to the charge of A but opposite sign since the grounding will provide electrons to the excess positive charge.

    I can see where I was going wrong on my original answer however I now believe this is the solution

    qinner = -Q1
    qouter = -Q2

    qnet = qinner + qouter = -Q1 - Q2

    Therefore a field will still exist between q and Q2 because Q2 is positively charged and the outer surface of q will be neutral thus inducing a -Q2 on the outer surface of q

    Thanks again
     
  10. Dec 16, 2011 #9

    Simon Bridge

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    Surely by that reasoning q(inner)=-Q1 and q(outer)=0 leaving a net charge on the middle sphere of -Q1?

    How did the -Q2 get there?
    The field due to the Q2 distribution anywhere inside it's radius is zero!
     
  11. Dec 17, 2011 #10
    I can't see how if Q2 is charged why a field will not exist inside it's radius that will be attracted to the middle sphere.
     
  12. Dec 17, 2011 #11

    Simon Bridge

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    Neither could Faraday (at first) - but he failed to detect one in experiments. It's the principle on which Faraday cages are built. Gauss' law shows you why not.
    In the problem - the center sphere is inside a Faraday cage.
     
  13. Dec 17, 2011 #12
    Of course! thanks for your help and patience.

    Regards
    Dan
     
  14. Dec 17, 2011 #13

    Simon Bridge

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    No worries - I'm amazed there are people still doing problems through December.
     
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