# Method of images tutorial?

I'm having an exam in Electrodynamics soon. I think I have most of it under control, but the method of images I'm not quite sure about.

There is not much in my book about, so I was thinking some of you might knew a website, article or something like that, maybe even with examples, that could help be get that last piece of the puzzle in my head?

I have tried Google, but I didn't find much to be honest. That's why I ask here.

-
Isn't it just a symmetry argument? – Bernhard Jun 18 '13 at 15:05
could you let us know what you can understand about it and what you're having problems with? – Jim Jun 18 '13 at 15:05
The main idea is, with a conductor with potential = 0, and a external charge $q$, to calculate the potential, in some domain (depending on the problem, this will be the whole space, or half-space, or interior of sphere, etc...), by adding an imaginary opposite charge $-q$ at some place, such as the sum of the 2 potentials created by the 2 charges will be zero on the conductor. – Trimok Jun 18 '13 at 16:43
The Method relies on the uniqueness of solutions to the boundary value problem for electric potentials. So you (1) remember the cases done in your book (planes and spheres typically) and (2) if you have to do a new geometry you must intuit the image distribution well enough to set up a solvable system to get the boundary potentials. – dmckee Jun 18 '13 at 17:01
I found the explanation in Volume II of the Feynman lectures to be quite illuminating. – Mark Allen Jun 18 '13 at 20:40

• An infinitesimally thin grounded sheet AB (I know the diagram is quite different but just assume it)
• A charge $q$ present at right side at distance $l$ from the sheet

"Let me tell you that our purpose of grounding is just to develop zero potential over the sheet. There is nothing more use of it."

So the Image Method let you assume that in this situation (i.e. In which there is zero potential sheet and there is a charge $q$ present at distance $l$ from it) then you can assume that there will be another charge of same magnitude as of $q$ but will be of opposite sign (i.e. $-q$ here).

I had tried to show that totally 'imaginary' image in the figure boxed inside.

Now the main point, You must be thinking that what is the use of it.

Well, from image method you can find the net force acting here which will be,

$$F=\frac{1}{4πε_0}.\frac{q.q}{(2l)^2}$$

which implies,

$$F=\frac{1}{4πε_0}.\frac{q^2}{4l^2}$$

Similarly you can find Electric field and other electrostatic terms.

-