1. Not finding help here? Sign up for a free 30min 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!

Potential due to induced charges

  1. May 28, 2013 #1
    1. The problem statement, all variables and given/known data
    A point charge q is placed at a distance ##r## from the centre O of a uncharged spherical shell of inner radius ##R## and outer radius ##2R##. The distance ##r<R##. Find the electric potential at the centre of shell.


    2. Relevant equations



    3. The attempt at a solution
    The charges will be induced as shown in attachment. The problem is calculating the potential. Do I have to consider each surface of both the spheres and calculate the potential? But that would give ##kq/r## which is wrong. :confused:
     

    Attached Files:

  2. jcsd
  3. May 28, 2013 #2

    TSny

    User Avatar
    Homework Helper
    Gold Member

    I don't quite understand your diagram. Looks to me like you have too many q's. There will be some charge induced on the inner surface of the shell and charge induced on the outer surface of the shell. Can you state the total amount of charge induced on each surface? Can you describe qualitatively how the charge will be distributed on each surface?
     
  4. May 28, 2013 #3
    [strike]What's wrong with the diagram? Won't the induced charges be equal to q in magnitude? [/strike] :confused:

    The q's near the surface are the induced ones. The one near the centre O is stated in the problem.

    Ah yes, the shells are not given to be conducting. Is my diagram correct for the case when the shells are connecting?

    But since the shells are not conducting, I don't know how the charge will be distributed. :(
     
  5. May 28, 2013 #4
    Hello Pranav

    Can you provide the answer to the problem ?
     
  6. May 28, 2013 #5
    As per the answer key, it is ##kQ(1/r-1/(2R))##.
     
  7. May 28, 2013 #6

    TSny

    User Avatar
    Homework Helper
    Gold Member

    The answer corresponds to a conducting shell. Note that there is only one shell. It has an inner radius R and an outer radius 2R.
     
  8. May 28, 2013 #7
    Sorry, I should have read the question properly. :redface:

    Now it is easy to answer if it is given that the shells are conducting, thank you! :smile:

    So the question is wrong as it isn't mentioned that the shells are conducting?

    And what if there were two separate shells?
     
  9. May 28, 2013 #8
    Okay...

    There are only two spherical surfaces not four as indicated by your diagram .

    The shell is assumed to be conducting in this problem.

    The charge induced on the inner surface(R) will be -q and that induced on the outer surface surface(2R) will be +q .

    Now replace the metal with two spherical charge distribution .One with radius R and charge -q and other with radius 2R and charge +q .

    Now in order to calculate potential at the center ,we have to find sum of potentials due to three charge distributions . +q(at distance r) , -q (sphere of charge of R radius ) ,+q(sphere of charge of 2R radius) .

    You very well know potential due to a point charge.

    What is the potential inside a thin spherical shell ?

    Plug in the values and you get the answer .

    Hope that helps .
     
  10. May 28, 2013 #9
    Thanks Tanya but you were a bit late, TSny cleared it up. :)
     
  11. May 28, 2013 #10

    TSny

    User Avatar
    Homework Helper
    Gold Member

    You have to be careful. Is the negative charge induced on the inner surface of the shell uniformly distributed? If not, how do you find the potential at the center from this non-uniformly distributed charge?
     
  12. May 28, 2013 #11
    If it is non-conducting, no, they are not uniformly distributed. The negative charge induced is equal in magnitude to q. The distance of the induced charges is at equal distance from O so is the potential again##-kq/R## due to the inner shell?
     
  13. May 28, 2013 #12

    TSny

    User Avatar
    Homework Helper
    Gold Member

    The induced charge will be uniform on the outer surface, but non-uniform on the inner surface.
     

    Attached Files:

  14. May 28, 2013 #13
    We are talking about the non-conducting case, right?
     
  15. May 28, 2013 #14

    TSny

    User Avatar
    Homework Helper
    Gold Member

    I was assuming a conducting shell. I think that's what leads to the answer you gave. If it's non-conducting, then I believe it would be fairly complicated. The induced charge would depend on the dielectric constant of the material and the surfaces of the shell would not be equipotential surfaces.
     
    Last edited: May 28, 2013
  16. May 28, 2013 #15
    Thanks TSny for the explanation. :smile:

    Can you give me a link which proves why the charges arrange like you have shown in your attachment? I did learn this a year ago but I seem to have forgotten now. :)
     
  17. May 28, 2013 #16

    TSny

    User Avatar
    Homework Helper
    Gold Member

    I don't know of a link offhand. One way to look at is as follows. There's a "uniqueness theorem" of electrostatics that states that if you can find any solution of your problem that satisfies all the boundary conditions, then it will be the only possible solution. Our boundary conditions here are that all points of the conductor must be at the same potential and the net charge of the conductor is zero. Also, the potential must go to zero at infinity.

    Consider the region of space outside the shell. This is a region of space bounded by the outer surface of the conductor and infinity. The outer surface of the conductor is an equipotential spherical surface and also the potential goes to zero at infinity. It can be shown that any solution to Laplace's equation for the electrical potential in this region with these boundary conditions must be spherically symmetric. (The potential must be of the form of the potential of a point charge located at the center of the shell.) This in turn requires the electric field outside the conductor to be spherically symmetric. Since the charge density at the outer surface of the conductor is proportional to the electric field at the surface, the charge density on the outer surface must be uniformly distributed over the surface.

    Note that this charge density on the outer surface produces no electric field inside the outer surface. So, it has no influence on how charge is distributed on the inner surface of the shell.

    Now consider the inner surface of the shell. The superposition of the electric field of the point charge inside the cavity and the induced charge on the inner surface must add to zero at every point inside the conducting material. If the point charge is not at the center of the shell, then you should be able to argue that the surface charge density on the inner surface cannot be uniformly distributed and still have zero net electric field inside the conducting material.
     
  18. May 28, 2013 #17
    Hello TSny

    Is the potential inside a spherical shell having uniform charge distribution and a shell having non uniform charge distribution same ?

    In this question , the charge distributed on the inside surface is not uniformly distributed .

    Is the potential inside still given by Kq/R ?
     
  19. May 28, 2013 #18

    TSny

    User Avatar
    Homework Helper
    Gold Member

    Hello, Tanya.
    For an arbitrary point inside the cavity, the potential for a charge Q distributed uniformly on the inner surface would not equal the potential at that point if Q is distributed non-uniformly. However, for the point at the center of the shell, you can show that the potential is kQ/R whether or not Q is distributed uniformly. [A bit of charge dQ on the inner surface produces the same potential at the center no matter where on the inner surface dQ is located. Therefore, V at the center will be the same no matter how the bits of charge dQ are distributed.]

    So, the answer you gave was correct and nicely stated.
     
    Last edited: May 28, 2013
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Potential due to induced charges
Loading...