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Method of Images (Electrostatics)

  1. Mar 21, 2009 #1
    1. The problem statement, all variables and given/known data

    A grounded spherical shell has inner radius of 1m. and outer radius of 2m.
    Q1=1 Coul. is located at x=0.5m and
    Q2=-2 Coul. is located at x=4m, respectively.

    2. Relevant equations

    • Find the charge system to find the potential outside the shell.
    • Find the charge system to find the potential inside the shell.
    • Obtain the expression for potential along the x axis.


    3. The attempt at a solution
    I found the image charge of Q1 outside the shell and Q2 inside the shell but i'm not sure about which surface i'm supposed to find the image charge of Q2 according to. Is it outer surface or inner one?

    And what is the difference between the charge system that we're using to obtain the potential of outside the shell and the charges we're using to find the potential inside the shell. Do they differ?
     
  2. jcsd
  3. Mar 21, 2009 #2

    gabbagabbahey

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    If the shell is a conductor, then both surfaces will be at the same potential.

    The difference is that you can't place image charges inside the reqion you are interested in (why?:wink:). So when you want to find the potential inside the shell, the image charges must be placed outside and vice versa.
     
    Last edited: Mar 21, 2009
  4. Mar 22, 2009 #3
    Is this because there should not be any extra charge at where the real charge stands.

    Another issue that i can not handle is
    How many image charges do i need to have?

    https://www.physicsforums.com/showthread.php?t=260107"

    Here you say, the boundary conditions which are zero potential for both inner and outer surface can not be satisfied with only one charge.

    But when i put two image charges outside the shell i have four unknowns:
    first image charge, second image charge and their distances to inner and outer surfaces.

    It seems so long and i'm totally confused :confused:
     
    Last edited by a moderator: Apr 24, 2017
  5. Mar 22, 2009 #4

    gabbagabbahey

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    Well, why is the method of images ever applicable in the first place? It has to do with a certain uniqueness theorem right? Are all the conditions of that theorem satsfied if there are image charges in the region of interest?

    You tell me. Like any physics problem, the best way to approach things is with a step by step process. Start by trying to find the image charge configuration needed to determine the potential outside the shell:

    (1)Where are you allowed to place image charges?
    (2)Where are the boundaries of the region you are interested in? What value do you want the potential to have on those boundaries?
    (3)If there were no conducting shell (and no image charges yet), what would the potential be on those boundaries?
    (4)Where could you place an image charge (and what magnitude would it need to be) so that the potential due to just Q1 and that image charge on those boundaries was zero?
    (5)Where could you place an image charge (and what magnitude would it need to be) so that the potential on those boundaries do to just Q2 and that image charge was zero?


    That was a very different problem; he needed to find the potential in the region N<r<M. Unlike your problem, that region was not filled with a conducting material and so the potential wasn't zero there. He had to place image charges in the regions r<N and r>M in such a way that would make V(N)=V(M)=0, and it turns out that the only way to do that requires an infinite number of charges.....your problem is much simpler!
     
    Last edited by a moderator: Apr 24, 2017
  6. Mar 22, 2009 #5
    (1) I am allowed to place the charges where i am not interested in.(For instance if i am interested in the Outer potential, i will place the image charge inside the shell)

    (2) For this question , let me interest in outside of shell, my boundary is r=2m. As the system is grounded, i want 0 potential at r=2m.

    (3) Thats one point i dont understand..Why would i need this potential? At where i am going to use it?

    (4) I don't think that i need Q1's image for outer shell. Because there shouldn't be any image charge where i am interested in.

    (5) Again for this question i found that 1C charge should be placed at x=1m.

    And in order to find out the potential of outside of system i need these charges: Q1 + Q2 + Q2's image(wrt the outer boundry)...
    If I were interested in the inside potential i would use Q2+ Q1 + Q1's image(wrt the inner boundry)

    If anystep goes wrong please correct me,
    And finally, what would happen if the system is ungrounded.. Need a hint again..
     
  7. Mar 22, 2009 #6

    gabbagabbahey

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    Well, you want the potential to be zero at r=2 right? If the potential were already zero there, you wouldn't need any image charges. For this reason, it's a good idea to check and see what the potential is there before blindly adding image charges...

    You can't place the image charge in the region you are interested in, but that doesn't mean there shouldn't be an image charge...

    And is this potential zero at the boundary?....If not, you've done something wrong :wink:
     
  8. Mar 22, 2009 #7
    You were right.. It wasn't zero potential at boundary according to my calculations.. but now i did something different but followed your advises.
    Still interested in the potential of outside the shell.
    Q1 + Q1i = 0 (for zero potential on r=2m shell) >> Q1i=-4C at x=8m;
    Q2 + Q2i = 0 (for zero potential on r=2m shell) >> Q2i=1C at x=1m;
    But i am going to use Q1,Q2,Q2i charge system to find the potential of outside shell, right?
    Again if anything goes wrong, correct me. Ty.
     
  9. Mar 22, 2009 #8

    gabbagabbahey

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    But that image charge is outside the shell; and you already found that you can't place image charges there since that is the region you are trying to find the potential in....

    I'll give you a hint: you are allowed to place an image charge at the same location as Q1:wink:

    That part is right.
     
  10. Mar 22, 2009 #9
    to satisfy the outer boundry conditions i will put a -Q1 as an image charge at the same location as Q1..

    That means on the outside of shell Q1 and its image Q1i (-Q1) will neutrlize each other and Q2 + Q2i will make the potential outside..

    If we were interested in potential inside we would put a -Q2 onto Q2 and Q1 + Q1i(wrt inner shell) would make the potential inside.. i got it now TY :)

    But now one more thing i wonder;
    what if it was ungrounded?
    Let there is a system with only Q1, there isn't any Q2 and ungrounded .. How should i start ?
     
  11. Mar 22, 2009 #10

    gabbagabbahey

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    Are you asking what the potential would be with just charge Q1 and an otherwise neutral conducting shell present?

    If so, just use Gauss' Law and the fact that E=0 inside a conductor. Determine the field first and then integrate it to find the potential.
     
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