Method of images: region of interest

[lys04]

Homework Statement

What does region of interest mean?

Relevant Equations

Poisson’s equation

I always thought region of interest is everything outside the conductor, which works when the image charges are placed inside the conductor. But sometimes the image charges are placed outside of the conductor instead which clearly changes Poisson’s equations since rho is now different?

 

[kuruman]

Say you have a spherical conducting shell of finite radius $R$ which is an equipotential. Put the origin at the center of the shell. Don't forget that you have another equipotential surface at infinity where the potential is usually defined to be zero. Now consider point P at $r<R$. Is it inside or outside the equipotential surfaces at $R$ and at infinity?

The region of interest is the region where the real charge is placed at point Q. Any other point P that can be reached from Q without crossing a conducting boundary is in the region of interest. So if Q is at $r<R$, the region of interest is $r<R$; if Q is at $r>R$, the region of interest is $r>R.$

 

[lys04]

Say you have a spherical conducting shell of finite radius $R$ which is an equipotential. Put the origin at the center of the shell. Don't forget that you have another equipotential surface at infinity where the potential is usually defined to be zero. Now consider point P at $r<R$. Is it inside or outside the equipotential surfaces at $R$ and at infinity?

The region of interest is the region where the real charge is placed at point Q. Any other point P that can be reached from Q without crossing a conducting boundary is in the region of interest. So if Q is at $r<R$, the region of interest is $r<R$; if Q is at $r>R$, the region of interest is $r>R.$

Ohh alright this makes sense. We were doing examples of having a point charge at x=a outside a sphere of radius R with constant potential and that the image charge should be placed at a position of x=R^2/a. This suggests that it’s possible for the image charge to be outside which is what made me confused because then wouldn’t rho in Poisson’s equation be different. But when the image charge is outside this means the real charge is inside the conductor so the region of interest is only r<R and Poisson’s equation isn’t changed.

 

[lys04]

Oh and by the Uniqueness theorem, the placement of image charges are unique too right?

 

[Orodruin]

Oh and by the Uniqueness theorem, the placement of image charges are unique too right?

No. The uniqueness theorem for the region of interest only involves the real charges. In order to invoke the uniqueness theorem for the images you would have to assume things about the solution outside the region in which the problem is to be solved.

For example, the way that the conductor actually compensates to create an equipotential surface is by rearranging a surface charge. Compare to solving a spherically symmetric scenario outside of some radius R. It doesn’t matter how the charge is distributed inside R, as long as the total charge is the same, the field outside R will be the same.

 

[kuruman]

Oh and by the Uniqueness theorem, the placement of image charges are unique too right?

You could reason this out yourself.

Suppose you place charge $q$ off-center inside a grounded conducting shell at $r<R.$ The potential inside the region of interest is the same as if there were no shell and image charge $q'$ is placed at $r'>R$ as specified by the method of images.

Now place real charge $q'$ at the same distance $r'>R$ as in the previous case and no charge at $r<R$. There will be an image charge $q''$ at $r''<R$ that makes the grounded shell an equipotential as specified by the method of images.

Would the image charge $q''$ and its position be the same as in the previous case, i.e. $q''=q$ and $r=r''$? Why or why not?