Method of images and spherical shell

In summary, to calculate the potential inside a spherical shell of outer radius M and inner radius N with a point charge Q at a distance r<M from center, the image charge must reside outside the shell.
  • #1
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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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  • #2
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).
 
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  • #3
gabbagabbahey said:
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).

why this is so?

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

Why must the image charge reside outside the shell?
 
  • #4
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.
 
  • #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]
 
  • #6
You might want to try putting all three charges on the z-axis instead (with z''>z'>b)
 
  • #7
ok, I'm not sure what the best approach is to solving this. Any hints?
 
  • #8
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]?
 
  • #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.
 
  • #10
gabbagabbahey said:
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).

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
 
  • #11
Where would you put the single image charge to make [itex]\phi(M)=0[/itex]? Is [itex]\phi(N)[/itex] zero with that configuration?
 

1. What is the method of images and how is it used in the study of spherical shells?

The method of images is a mathematical technique used to solve problems involving electric charges near conductive surfaces. In the study of spherical shells, it is used to calculate the electric field and potential at points outside and inside the shell.

2. What is the principle of superposition and how is it applied in the method of images?

The principle of superposition states that the combined effect of multiple sources is equal to the sum of the individual effects. In the method of images, this means that the electric field and potential at a point is the sum of the contributions from the real charge and its corresponding image charge.

3. How is the position and magnitude of the image charge determined in the method of images?

The position and magnitude of the image charge are determined using the boundary conditions at the surface of the spherical shell. The image charge is placed at a point that satisfies these conditions and its magnitude is chosen to cancel out the electric field produced by the real charge at the surface.

4. Can the method of images be used for non-spherical shells?

Yes, the method of images can also be applied to problems involving non-spherical shells. However, the calculation of the image charge becomes more complex and often requires the use of numerical methods.

5. What are some applications of the method of images in physics?

The method of images is commonly used in electrostatics and electromagnetism to solve problems involving conductive surfaces, such as capacitors, conducting spheres, and charged cylinders. It is also applied in other fields, such as fluid mechanics and heat transfer, to study the behavior of fluids and heat flow near boundaries.

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