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Method of images

  1. Sep 28, 2008 #1
    A grounded & conducting spherical shell of outer radius M and inner radius N has a point charge Q inside the shell at a distance r<M from center. Using method of images, find the potential inside the sphere.

    Could I just use the superposition of charge and point charge?

    [tex] \phi(r) = \frac{1}{4 \pi \epsilon _0} \left( \frac{q}{|r-r'|} + \frac{q'}{|r-r''|} \right) [/tex]
     
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  3. Sep 28, 2008 #2

    gabbagabbahey

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    You know that since the shell is a grounded conductor, [itex]\phi(M)=\phi(N)=0[/itex]. You cannot achieve that with only one extra charge. You will need two extra charges; q' and q'' outside the shell (r>M).
     
    Last edited: Sep 29, 2008
  4. Sep 29, 2008 #3
    why this is so?

    Why would this change for a conductor that was not grounded?

    Why must the image charge reside outside the shell?
     
  5. Sep 29, 2008 #4

    gabbagabbahey

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    Well, the surfaces of a conductor are equipotentials; so if the conductor wasn't grounded you would have [itex]\phi (N)=\phi _N[/itex], [itex]\phi (M)=\phi _M[/itex] where [itex]\phi _N[/itex] and [itex]\phi _M[/itex] are constants. Since the conductor is grounded these will both be zero.

    In either case you need an image charge configuration that will encompass 4 things:

    (1)Ensure that the potential at r=M is a constant [itex]\phi _M[/itex] (zero for a grounded conductor)
    (2)Ensure that the potential at r=N is a constant [itex]\phi _N[/itex] (zero for a grounded conductor)
    (3)Ensure that the only charge present in the region N<r<M is your actual point charge Q
    (4)Ensure that there are no additional charges in the region you are calculating phi for (r<N)

    (3) and (4) mean that any image charges will have too be placed at r>M. While, (1) and (2) each give you a constraint that requires an image point charge. You probably know that you can get a constant potential for a single spherical surface by placing a single image charge outside the surface; well by the superposition principle you can therefor achieve a constant potential on both surfaces by placing two charges at convenient locations.
     
  6. Sep 29, 2008 #5
    Here's what I'm trying, let me know if this is okay:

    Using CGS units:

    [tex] \phi(\vec{r}) = \frac{q}{|\vec{r}-\vec{x}|}+ \frac{q'}{|\vec{r}-\vec{y}|}+\frac{q''}{|\vec{r}-\vec{z}|}[/tex]

    [tex] \phi(a) = \frac{q/a}{\left| \hat{r}-\frac{x}{a}\hat{x} \right|}+ \frac{q'/a}{\left| \hat{r}-\frac{y}{a}\hat{y} \right|} +\frac{q''/a}{\left| \hat{r}-\frac{z}{a}\hat{z} \right|}=0[/tex]

    [tex] \phi(b) = \frac{q/b}{\left| \hat{r}-\frac{x}{b}\hat{x} \right|}+ \frac{q'/b}{\left| \hat{r}-\frac{y}{b}\hat{y} \right|} +\frac{q''/b}{\left| \hat{r}-\frac{z}{b}\hat{z} \right|}=0[/tex]
     
  7. Sep 30, 2008 #6

    gabbagabbahey

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    You might want to try putting all three charges on the z-axis instead (with z''>z'>b)
     
  8. Oct 1, 2008 #7
    ok, I'm not sure what the best approach is to solving this. Any hints?
     
  9. Oct 1, 2008 #8

    gabbagabbahey

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    hmmm... maybe start with an easier problem; forget the second image charge for a minute; where would you have to put an image charge (and what would the charge it have to be) to make the potential on the inner surface zero? How about a constant [tex]V_1[/tex]?
     
  10. Oct 1, 2008 #9
    I've read over the examples in the book and they seem to do something along the lines of:


    [tex]\frac{q/a}{\left| \hat{r}-\frac{z}{a}\hat{z} \right|} = - \frac{q'/a}{\left| \hat{r}-\frac{z'}{a}\hat{z} \right|} -\frac{q''/a}{\left| \hat{r}-\frac{z''}{a}\hat{z} \right|}[/tex]

    [tex]\frac{q/b}{\left| \hat{r}-\frac{z}{b}\hat{z} \right|} = - \frac{q'/b}{\left| \hat{r}-\frac{z'}{b}\hat{z} \right|} -\frac{q''/b}{\left| \hat{r}-\frac{z''}{b}\hat{z} \right|}[/tex]

    the problem is I can simply continue along side of the book where they would continue as such:

    [tex]\frac{q}{a}=-\frac{q'}{a}-\frac{q''}{a}[/tex]

    [tex]\frac{z}{a}=\frac{z'}{a}+\frac{z''}{a}[/tex]

    and similarly for r=b

    This is where I am stuck... not sure how to get the relations since the case is not as simple.
     
  11. Oct 2, 2008 #10
    I think that only one charge is needed since the outside shell has no idea what the inside shell is doing since the conductor is grounded and there is no e-field to transmit the information from the inside shell to the outside shell
     
  12. Oct 2, 2008 #11

    gabbagabbahey

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    Where would you put the single image charge to make [itex]\phi(M)=0[/itex]? Is [itex]\phi(N)[/itex] zero with that configuration?
     
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