
#1
Sep2808, 10:21 PM

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<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}{rr'} + \frac{q'}{rr''} \right) [/tex] 



#2
Sep2808, 11:01 PM

HW Helper
P: 5,004

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).




#3
Sep2908, 08:28 PM

P: 148

Why would this change for a conductor that was not grounded? Why must the image charge reside outside the shell? 



#4
Sep2908, 09:32 PM

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 [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
Sep2908, 11:45 PM

P: 148

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
Sep3008, 11:00 PM

HW Helper
P: 5,004

You might want to try putting all three charges on the zaxis instead (with z''>z'>b)




#7
Oct108, 12:34 AM

P: 148

ok, I'm not sure what the best approach is to solving this. Any hints?




#8
Oct108, 12:38 AM

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 [tex]V_1[/tex]?




#9
Oct108, 08:48 AM

P: 148

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
Oct208, 01:49 PM

P: 148




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