- #1

fluidistic

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## Homework Statement

I would like to

1)Find the electric field in all the space in the following set up: There's an electric charge q in a region with dielectric constant ##\varepsilon _1## which is at a distance d from a plane that separates the space into 2 regions, namely the one of the charge and another region with dielectric constant ##\varepsilon _2##.

2)Find the polarization charge density over the plane. (Jackson gives out the answer to that task though he does not give the details which is awesome for me since I have to train hard).

## Homework Equations

To get the electric field in all the space I will use the method of images (just like Jackson's do in order to get ##\Phi## in all the space)

Page 155 and 156 in the 3rd edition of Jackson's book.

The main idea I believe is first to get the potential in all the space and then get ##\vec E## via ##\vec E = \nabla \Phi##.

## The Attempt at a Solution

I followed Jackson's book up to where he found out the potential in the region of the electric charge. It's ##\Phi (\rho, \phi , z )= \frac{1}{4\pi \varepsilon _1 } \left [ \frac{q}{\sqrt{\rho ^2 + (d-z)^2}} + \frac {q'}{\sqrt{\rho ^2 + (d+z)^2}} \right ]## where q' is a pseudo-charge situated in the region with dielectric constant ##\varepsilon _2## at a distance d from the plane (well it's the conventional method of images).

And then I'm lost when he derives the potential in the region with ##\varepsilon _2##.

He says that since there's no charge in that region, the potential must be a solution to Laplace equation without singularities. So far so good. It's just ##\varepsilon _2 \nabla \cdot \vec E =0##, just as he himself wrote a few lines before.

But then he says that "Clearly the simplest assumption is that for z<0 the potential is equivalent to that of a charge q'' at the position A of the actual charge q: ##\Phi = \frac{q''}{4\pi \varepsilon _2 \sqrt{\rho ^2 + (d-z)^2}}##, for z<0.""

???

I don't understand why he isn't summing up the potential due to q and q' like in the first region. He now uses another pseudo charge which is located at the same spot as the real charge. It's like he's not using the method of images anymore but he somehow thinks the problem as an "effective charge" right on the spot of the real charge. I don't understand why this work. I know that since the solution to Laplace equation is unique here, if this method works then it works and it leads to the only solution but I just don't understand the strategy here.

Thanks for any clarification.