# method of images

by LocationX
Tags: images, method
 P: 148 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
 HW Helper P: 5,004 You know that since the shell is a grounded conductor, $\phi(M)=\phi(N)=0$. You cannot achieve that with only one extra charge. You will need two extra charges; q' and q'' outside the shell (r>M).
P: 148
 Quote by gabbagabbahey You know that since the shell is a grounded conductor, $\phi(M)=\phi(N)=0$. 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?

HW Helper
P: 5,004

## method of images

Well, the surfaces of a conductor are equipotentials; so if the conductor wasn't grounded you would have $\phi (N)=\phi _N$, $\phi (M)=\phi _M$ where $\phi _N$ and $\phi _M$ 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 $\phi _M$ (zero for a grounded conductor)
(2)Ensure that the potential at r=N is a constant $\phi _N$ (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.
 P: 148 Here's what I'm trying, let me know if this is okay: Using CGS units: $$\phi(\vec{r}) = \frac{q}{|\vec{r}-\vec{x}|}+ \frac{q'}{|\vec{r}-\vec{y}|}+\frac{q''}{|\vec{r}-\vec{z}|}$$ $$\phi(a) = \frac{q/a}{\left| \hat{r}-\frac{x}{a}\hat{x} \right|}+ \frac{q'/a}{\left| \hat{r}-\frac{y}{a}\hat{y} \right|} +\frac{q''/a}{\left| \hat{r}-\frac{z}{a}\hat{z} \right|}=0$$ $$\phi(b) = \frac{q/b}{\left| \hat{r}-\frac{x}{b}\hat{x} \right|}+ \frac{q'/b}{\left| \hat{r}-\frac{y}{b}\hat{y} \right|} +\frac{q''/b}{\left| \hat{r}-\frac{z}{b}\hat{z} \right|}=0$$
 HW Helper P: 5,004 You might want to try putting all three charges on the z-axis instead (with z''>z'>b)
 P: 148 ok, I'm not sure what the best approach is to solving this. Any hints?
 HW Helper P: 5,004 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 $$V_1$$?
 P: 148 I've read over the examples in the book and they seem to do something along the lines of: $$\frac{q/a}{\left| \hat{r}-\frac{z}{a}\hat{z} \right|} = - \frac{q'/a}{\left| \hat{r}-\frac{z'}{a}\hat{z} \right|} -\frac{q''/a}{\left| \hat{r}-\frac{z''}{a}\hat{z} \right|}$$ $$\frac{q/b}{\left| \hat{r}-\frac{z}{b}\hat{z} \right|} = - \frac{q'/b}{\left| \hat{r}-\frac{z'}{b}\hat{z} \right|} -\frac{q''/b}{\left| \hat{r}-\frac{z''}{b}\hat{z} \right|}$$ the problem is I can simply continue along side of the book where they would continue as such: $$\frac{q}{a}=-\frac{q'}{a}-\frac{q''}{a}$$ $$\frac{z}{a}=\frac{z'}{a}+\frac{z''}{a}$$ 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.
P: 148
 Quote by gabbagabbahey You know that since the shell is a grounded conductor, $\phi(M)=\phi(N)=0$. 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
 HW Helper P: 5,004 Where would you put the single image charge to make $\phi(M)=0$? Is $\phi(N)$ zero with that configuration?

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