Why Is the Potential Different for Dielectric Constants in the Method of Images?

[Lindsayyyy]

Hi everyone

I have an understanding problem concerning the method of images method

Homework Statement

Jackson uses this kind of problem in his textbook but I don't understand it.
We have a charge q on the z axis at distance d. the dielectric constant is e1 when z>=0 and e2 when z<0. Now I want to find the potential.

Homework Equations

The Attempt at a Solution

This is what Jackson does what I don't understand.

We use another charge q' at the opposite location to get the potential for z>0

\Phi =\frac {1}{4 \pi \epsilon_1} (\frac{q}{R_1}+\frac{q&#039;}{R_2})

This is equal to the case where we have only a conducting plate on the x axis. Now we have to find the potential for z<0 whereas Jackson says:
\Phi =\frac {1}{4 \pi \epsilon_2} \frac {q&#039;&#039;}{R_1}

I don't understand this step. The explantion which is given from Jackson doesn't help me either. How to I get to this solution?

Thanks for your help.

 

[TSny]

Now we have to find the potential for z<0 whereas Jackson says:
\Phi =\frac {1}{4 \pi \epsilon_2} \frac {q&#039;&#039;}{R_1}

I don't understand this step. The explantion which is given from Jackson doesn't help me either. How to I get to this solution?

Jackson is making an "educated guess" or "ansatz" and then showing that everything works out for a certain value of q''.

This guess can be motivated by a couple of observations:

(1) For determining the potential in the region z < 0, any image charge(s) must be placed in the region z > 0.

(2) The image charge(s) used to produce the potential for z < 0 must produce a potential that satisfies the boundary conditions at every point on the z = 0 plane (Jackson's 4.42 in second edition).

These two conditions pretty much require the image charge q'' to be placed in region z > 0 at the same location as the real charge q.

 

[Lindsayyyy]

Thanks for your reply, but I actually still don't understand the procedure. It feels like you have to know the solution in order to solve it. There is no logical approach for me :(
Why do I have to put a charge in z>0 in order to determine the potential in z<0?

 

[TSny]

In this problem it should be clear that the solution for $\Phi(x, y, z)$ will be axially symmetric about the z-axis in both regions. Thus, any image charges will probably have to be placed on the z-axis to preserve this symmetry.

$\Phi$ will have to satisfy Laplace's equations at all points except the point where the real charge q is located. (This follows from equations (4.41) 2nd ed). Thus, in region z>0, $\Phi$ cannot have any singularities except at the location of q. So, when setting up images to produce the solution in the region z>0, the image charges must be placed on the z-axis in the region z<0. It is not at all obvious to me that the solution in region z>0 can be obtained by using just one image charge q' located in region z<0 at the same distance from z = 0 as q. But, it does turn out to work.

There can be no singularities for $\Phi$ in region z<0 since there are no real charges in this region. So, when deciding on where to place image charges for producing $\Phi$ in region z<0, the image charge(s) will need to be placed on the z axis in the region z>0. Using the observations that I gave in my previous post, it does seem natural to me to try a single image charge q'' located at the same position as q. The values of q' and q'' are then determined from the boundary conditions.

For me, there is an element of guessing in deciding where to place the images. (Sort of like learning integration techniques. In a complicated integral, you're not sure if a certain trig substitution, say, is going to work until you give it a try.) To my knowledge, there are only a relatively few geometries where you can use image charges and you get familiar with them by just studying solutions that are given in standard texts.

 

[Lindsayyyy]

ok thanks for the help. if z>0 is in vacuum, the problem wouldn't change, would it? Except for e1=1 then.

 

[TSny]

That's correct.