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!

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]
  2. jcsd
  3. Sep 28, 2008 #2


    User Avatar
    Homework Helper
    Gold Member

    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


    User Avatar
    Homework Helper
    Gold Member

    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


    User Avatar
    Homework Helper
    Gold Member

    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


    User Avatar
    Homework Helper
    Gold Member

    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:



    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


    User Avatar
    Homework Helper
    Gold Member

    Where would you put the single image charge to make [itex]\phi(M)=0[/itex]? Is [itex]\phi(N)[/itex] zero with that configuration?
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Method of images
  1. Method of images (Replies: 7)

  2. Method of images (Replies: 1)

  3. Method of images (Replies: 3)

  4. Method of images (Replies: 3)

  5. Method of Images (Replies: 2)