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Method of images

by LocationX
Tags: images, method
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LocationX
#1
Sep28-08, 10:21 PM
P: 147
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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gabbagabbahey
#2
Sep28-08, 11:01 PM
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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).
LocationX
#3
Sep29-08, 08:28 PM
P: 147
Quote Quote by gabbagabbahey View Post
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?

gabbagabbahey
#4
Sep29-08, 09:32 PM
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Method of images

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.
LocationX
#5
Sep29-08, 11:45 PM
P: 147
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]
gabbagabbahey
#6
Sep30-08, 11:00 PM
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You might want to try putting all three charges on the z-axis instead (with z''>z'>b)
LocationX
#7
Oct1-08, 12:34 AM
P: 147
ok, I'm not sure what the best approach is to solving this. Any hints?
gabbagabbahey
#8
Oct1-08, 12:38 AM
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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]?
LocationX
#9
Oct1-08, 08:48 AM
P: 147
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.
LocationX
#10
Oct2-08, 01:49 PM
P: 147
Quote Quote by gabbagabbahey View Post
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
gabbagabbahey
#11
Oct2-08, 07:36 PM
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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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