1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Charge Between Two Conducting, Connected Shells

  1. Dec 27, 2012 #1
    1. The problem statement, all variables and given/known data
    On the XY plane there exist two concentric, thin, conducting shells, centered at the origin; one is of radius R, the other of 2R.
    They are connected.
    There exists a point charge at (0, 1.5R).
    What is the potential at some point outside the outer ring?

    2. Relevant equations
    I don't really understand what's going on here

    3. The attempt at a solution
    Tried method of image charges for outer sphere, and inverse for inner sphere... didn't get anywhere.
    I'm trying to understand the general behavior of charges inside a void inside a conductor...
    I was told that given any conductor, with a void inside of it containing charge, the electric field outside of it will be the same irrespective of how you move the charge around..
    I'd really appreciate an explanation!
  2. jcsd
  3. Dec 28, 2012 #2


    User Avatar
    Homework Helper

    The question is about the electric field outside the rings.

    The charge in the void attracts opposite charges to the inner surfaces of the conducting shells, which result in the same surface charge on the outer surface. But the electric field in the void does not penetrate through the conducting wall of the outer shell. The charges on the outer surface "do not know" about the inner electric field. The outer electric field is the same as that of a charged sphere of radius 2R.

    There is also a law of Electrostatics that you can fill a closed equipotential surface with metal, without changing the field outside.
    The two shells are connected, so they are at the same potential. You can fill the void with metal. What happens with the charge inside?

  4. Dec 28, 2012 #3
    How is it that the charge attracts opposite charges to the inner surfaces of the conductors, if it is outside of one of them?
  5. Dec 28, 2012 #4


    User Avatar
    Homework Helper
    Gold Member
    2017 Award

    Just for clarification, are we dealing with "rings" or "spheres" (or perhaps even cylinders perpendicular to the xy plane)?
  6. Dec 28, 2012 #5
    Spherical shells, or, just for 2D, rings. Both very thin
  7. Dec 28, 2012 #6


    User Avatar
    Homework Helper

    I meant "inner" with respect to the void.

  8. Dec 28, 2012 #7
    I got it. It's essentially no different from a charge in a void of a conductor.
    There can be no field between the shells, so there is a total charge distribution equal to the point charge inside on the outer shell. Using the '4πσ argument', the distribution on the outer shell must be uniform since there are no charges outside. Therefore the field outside is like that of a point charge in the center of the shells.
  9. Dec 28, 2012 #8


    User Avatar
    Homework Helper

    There is field between the shells in the void, but that does not influence the outer field. Yes, the field outside is the same as of a point charge in the centre.

  10. Dec 28, 2012 #9
    There is field? I don't understand why..
  11. Dec 28, 2012 #10


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    The charge on the point charge will be opposite in sign to that on the nearest parts of the conductor, so not only is there a field in the cavity, it's stronger than with the point charge alone.
  12. Dec 28, 2012 #11
    But the shells are connected, don't they become one conductor and make the field in the void between them zero?
    Could you explain this in more depth please?
  13. Dec 28, 2012 #12


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    They can't make the field in the void zero if there are point charges in it. Imagine getting really close to one of those point charges. As you get closer, the potential due to the point charge increases without limit, while that due to the relatively distant conductor is barely changing. So net potential changes, so there is a field.
  14. Dec 28, 2012 #13
    Gotcha, that makes sense. So there is field. Can you please elaborate more as to why exactly the solution remains the same? Why does this field not affect anything outside? What exactly is going on in this problem?
  15. Dec 28, 2012 #14
    I just know for certain that the field outside of the conductor behaves like that of a point charge at the origin. I know this must be true irrespective of what's going on in the conductor by guessing a solution to Laplace's Equation with the boundary conditions of a shell-like non-zero equipotential surface, and a zero equipotential surface at infinity. I can't even justify why the point charge at the at the origin should be of same charge q, because I don't know the potential on the conductor..
    I'd really appreciate a detailed explanation
  16. Dec 28, 2012 #15
    The problem is phrased with shells that have zero width, which really made things difficult to understand. I'm picturing them as having width, since zero width doesn't exist in nature. Then I looked at the system without the connection nor the inner conducting shell. In that instance it was clear to me why the same charge of opposite sign moves to the inner boundary of the outer shell - applying gauss's law and requiring there be no field inside the outer shell. Conservation of charge means the exact same charge as the point charge is spread across the outmost boundary of the outer shell. That helped, because the null field inside the outer shell is exactly what prevents the very outmost charges from feeling and force/being aware of electric field from inside. This is why they distribute uniformally and therefore field outside is that of a point charge in the center.
    Would still very much like to know whether the inner conducting shell affects the problem in any way, and also, from guessing a solution to Laplace's Equation, how could one determine that the magnitude of the virtual point charge creating the field is the same of the original point charge in the problem..
  17. Dec 28, 2012 #16


    User Avatar
    Homework Helper

    Maybe we do not understand the same thing on the "inside of the outer shell". The shall has a wall of some thickness, although it is negligible compared to the radius. The electric field is zero inside the wall, but there is an electric field around the charge in the void between the shells. Think to make a Gaussian surface around the charge Q. The surface integral for this surface can not be zero as it encloses charge. At the same time, the electric field is zero inside the inner shell, as it does not contain any charge.
    The electric field lines emerge from charges and end in charges or at infinity. There must be some surface charge on the inner surface of the big shell and also on the outer surface of the small shell. The shells are connected, they are at the same potential, and the charges can move from one shell to the other. The charge Q induces -Q surface charge on the conductors at the interface between conductor and void. The conductors are electrically neutral, so Q charge has to appear at the outer surface. (As the electric field is zero inside the small shell, there is no surface charge on its inner surface )


    Attached Files:

    Last edited: Dec 28, 2012
  18. Dec 28, 2012 #17
    Thank you very much! That was really clear. I came pretty close in my last post as you see.. Definitely not as clear cut though. A question remains - if I were just guessing a solution to Laplace's Equation - how would I know what magnitude to assign the charge at the origin?
    I'm asking because whenever there's a problem area whose outer boundary is a spherically symmetric conductor, a point charge in its center solves Laplace's Equation for the area outside the conductor.. I just need to know what magnitude of charge to assign it..
  19. Dec 28, 2012 #18


    User Avatar
    Homework Helper

    You want the solution of the Laplace equation in the space between a spherical surface and infinity, with the boundary condition that the potential is constant on the spherical surface and zero at infinity.
    The solution of the Laplace equation with given boundary condition is unique. If you find a solution, it is the solution.

    The potential around a single conducting sphere is spherically symmetric. If the spherical surface encloses Q charge, the electric field is kQ/r^2 according to Gauss' Law, and the potential is kQ/r at distance r from the centre if r>R.

    In your problem, the outer spherical surface encloses the charge Q placed into the void. So the potential is kQ/r for r>R.
    To find the electric field and potential inside the cavity would be more difficult, but it was not the question.

  20. Dec 29, 2012 #19
    I found this thread rather helpful in addressing a similar difficulty I had. Can someone verify I understand this problem:

    Given a system of two concentric spherical shells, these are the areas I can divide it into:

    1) Outside both shells.
    2) The outer layer of the outer shell
    3) The inner layer of the outer shell (or equivalently, the thickness below the outer layer if thinking microscopically)
    4) The void between the shells.
    5) The outer layer of the inner shell
    6) The inner layer of the inner shell
    7) The area below the inner layer of the inner shell.

    There exists a point charge Q at 1.5R.

    So, this is what happens:

    1) For r > 2R, the electric field behaves as a point charge Q at the center. This is because the electric field from the point charge, even though its spherically asymmetric, is canceled out by the inner surface of the outer layer. Therefore, the Gaussian surface I construct has to only worry about the surface charge and my E dot dA integral reduces to a constant.
    2) The outer surface of outer layer has a charge Q distributed over it, spherically symmetric.
    3) ...because the inner surface of the outer layer has charge -Q distributed so as to cancel out the electric field lines (within the conductor itself) from point charge at 1.5R as this is the behavior of any conductor.
    4) In the void, I'm not sure exactly what's happening because the point charge isn't exactly spherically symmetric but I know there's some sort of electric field. In fact, if I were to just consider equipotential surfaces for the point charge without the influence of anything else, there are some surfaces that actually intersect the outside conductor at some points but don't at others. Seems really weird but is my interpretation correct?
    5) So at this point... does the outer surface of the inner shell also have a charge of -Q distributed over it to cancel out the electric field lines such that the conductor's thickness has an electric field of zero (and the inner surface gains charge +Q)? It seems strange that there is -Q distributed in two places to cancel out the electric field from the point charge.
    6) Applying Gauss law now on the inside, the Electric field is zero.

    However, I'm slightly confused as to the case in which the a cavity DOES have an electric field of zero. So electrostatic shielding only applies when:

    1) There are no point charges in conductor cavities
    2) Your within the conductor itself when point charges are present in cavities.

    Also, given the spherical asymmetric of the point charge, why can I assume the charge Q is distributed uniformly? Again, is this because the electric field is cancelled out by the inner surface so the outer surface uniformly distributes itself?
    Last edited: Dec 29, 2012
  21. Dec 30, 2012 #20


    User Avatar
    Homework Helper

    Yes. There is a surface charge Q on the outer surface of the big shell. These charges do not feel other electric force but that from each other, so distribute themselves symmetrically. As the surface charge is constant on the sphere, the electric field is spherically symmetric, independent on the angles. Your Gaussian integral for a concentric sphere and radius r >2R reduces to 4πr2E=Q/ε0.

    In the problem, the shells are connected with a wire. Both shells are at the same potential. There are negative surface charges induced also on the outer surface of the inner shell, and the sum of the induced charges on both surfaces is equal to Q(in)+Q(out)= -Q.
    If the shells are insulated I think, there would be -Q charge induced on the inner surface of the outer shell and no induced charge on the inner shell.

    If there is a single point charge without anything else the equipotential surfaces are spherical and concentric with the point charge. But the point charge is between two equipotential surfaces, as a metal surface is always equipotential. In the problem, both surfaces are at the same potential. The field lines are normal to the equipotential surfaces, so they are normal to the surfaces of both shells very near to them. The electric field lines emerge from charges and end in charges. So the electric field in the cavity at a point P of the surface is σ(P)/ε0, where σ is the surface charge density at that point.

    No, the surface charges on both shells on the inner surface of the cavity add up to give -Q. As the shells are connected, there can be a negative net charge on the inner shell -q=Qin, and q net charge on the outer shell, so Qout=-Q+q, so the the surface charge on the outer surface of the outer shell is q-Qout=Q.

    That is true. No concentric spherical surface has any charge in the void inside the inner shell.

    Electric shielding applies against outer electric fields: If you place charges outside the shells at r>2R the electric field would not change in the void.

    You get the electric field by solving Laplace equation and appliying the boundary conditions. But sometimes you can calculate it by assuming charged surfaces substituting metal plates/shells. Do not forget that the electric field lines can not cross really the metal layer. The electrons on the outer surface do not feel the inner charge.

Share this great discussion with others via Reddit, Google+, Twitter, or Facebook