A dielectric + an infinite line of charge = use method of images

[redglasses]

Hi everyone, first time poster, long time reader (only because you're all so clever and I feel my contribution wouldn't be all that helpful haha).

Homework Statement

A dielectric (homogeneous, isotropic, with relative permativity \epsilon_{r}) is located in the region x<0.

An infinite line of charge \lambda, parallel to surface of dielectric slab and parallel to z axis, is located at (x,y,z) = (d,0,0). x>0 is a vacuum.

We are told that the field in the region x>0 can be found by way of the method of images, by introducing an image charge \lambda^{&#039;}, at (-d,0,0); and the field in the region x<0 can be found by replacing \lambda with \lambda^{&#039;&#039;}.

Homework Equations

E = -\nablaV
Surface charge: \sigma = \epsilon_{0}*E \cdot \widehat{n}

The Attempt at a Solution

This problem is confusing me. Up until now it was my understanding that I could only really use the method of images when at the boundary the potential is equal to zero. That being said I did see an example in class where the potential was constant, but I found this rather confusing.

What I tried to do for the first part of the question, more as a thought process than by way of calculation (as for some reason I keep getting stuck finding the potential as I'm far more used to working more in 2 dimensions with image charges instead of lines of charge for this kind of problem) is to try to set up a similar problem where at the boundary it is an infinite grounded conducting plane (ie V = 0), try to find the surface charge that would be induced by \lambda and integrating to find \lambda^{&#039;} and showing they're equivalent.

But by my luck that's probably the completely wrong approach! I have a feeling I've misunderstood a crucial concept.

I understand this is a poor attempt at a solution - I have almost resigned myself to the fact that I won't nut out this question without seeing my lecturer, but any small pushes in the right direction would be gladly accepted and greatly appreciated!

 

[gomboc]

Unfortunately, the method of image charges gets a little more advanced when you are dealing with a dielectric rather than a conductor. Most image questions involve an infinitely thin conducting sheet, because a for a "perfect conductor" this is sufficient. They could use an infinitely thick conductor though if they wanted.

With a dielectric, it needs to have some thickness, since it has resistance. Fortunately an infinitely thick dielectric is the simplest case (except of course for an infinitely thin one, in which case it could just be neglected!).

Given your line charge at (d,0,0), you will want to model your dielectric differently depending on which side of the boundary you're on. For x>0, outside the dielectric, you need to add an image line charge to mimic the dielectric at (-a,0,0), not necessarily the same distance from the boundary as would be the case for a conductor.

But as you said, you don't have zero potential at the boundary now. Even though the question doesn't ask it, you need to find the potential inside the dielectric is well. You do this by adding an image line charge at (b,0,0) - i.e. there are two separate "cases" needed to solve the problem. (both image charges are assumed to be of unknown \lambda

Write up two equations for the potential then - one for x>0 and one for x<0.

Here's where the trick comes in - rather than knowing that V must be 0 at the boundary, we say that it must be continuous at the boundary. Using this condition, you can set the two potential equations equal at x=0.

You need another boundary condition though to solve for all the variables. That is that the electric field must also be continuous at the boundary. This means that your other boundary condition is

\epsilon_r \left(\frac{\partial V_{x&lt;0}}{\partial x}\right)_{x=0} = \left(\frac{\partial V_{x&gt;0}}{\partial x}\right)_{x=0}

That's the key. Then it's just a matter of mathematics solving the resulting systems of equations for the two image line charges positions and magnitudes.

NOTE that one result of solving these equations will be that a=b=d, meaning that one of the image charges (b,0,0) is actually right on top of your original line charge, and thus amounts to changing the magnitude of the original line charge as the question states.

In a nutshell, this problem just makes clear how the method of images really works - applying the first boundary condition is really obvious with an infinite conducting sheet, and appyling the second is unneccesary in many cases.

Hope this helps!

 

[redglasses]

Thank you gomboc for your long and helpful reply!
That really improved my understanding of the overall problem. It's only the little (and probably simpler :P) steps of the problem that I'm having some trouble with at the moment, but I'll give it another good bash now.