Electromagnetics - Method of Images ?

1. Jul 11, 2006

s0undmind

Electromagnetics - Method of Images ???

Hello all, I am in the process of preparing for a entrance examination. In the course of studying I came across this problem that I just can not figure out. Here it is:

A perfectly conducting plane is located in free space at x=4, and a uniform inifinte line charge of 40nC/m lies along the line x=6, y=3 (the intersection of these planes). Let V=0 at the conducting plane. At P(7,-1,5) find (a) V (potential at P); (b) E (electric field at P).

Answer: -316V ; -45.4ax V/m (ax is the unit vector in the x direction)

I have the answers as you can see but I have no idea on how to arrive to them. I have tried for hours. I would appreciate any help or comments.
THank you.

2. Jul 12, 2006

Staff: Mentor

Are you sure about the answer to (b)? It sure seems like there would be a y component to the E-field at (7,-1,5).

What equation do you use to calculate the Electric field due to a charge distribution? Once you have the E field, how do you calculate the voltage V?

3. Jul 15, 2006

s0undmind

Thank you for your reply. Well, I started to think about it again and I agree that there should be a y component. I guess it is a typo in the text especially since I was able to calculate a 45.4ax component (corresponding to the egative of x component in the answer) and a 99ay component. I added the electric field contributions from the uniform charge density(at x=6,y=3,z) and its image (at x=2,y=3,z) and obtained this answer.

For the Voltage, I performed an ugly line integral from a point on the conducting surface. I chose a point lying on a normal line from the conducting surface to P(7,-1,5) so that only x would vary in the integral.
After some manipulations and using the wolfram integrator I arrived at the answer of 343.5V, which is close but does not agree with the answer in the text and the sign is wrong. I wonder if there's an easier way to solve this problem. If you have any hints I would like to hear to them.

Thank you.

4. Aug 25, 2006

leright

I know this thread is old, but I would like to say that this question is poorly explained. A voltage is measured between two points. A voltage is simply a path integral of the E-field between two points (for electrostatic conditions the path taken to get from one point to the other does not matter since the field is conservative...the only thing that matters is the point you start at and the point you end at....YES, this is important). The problem, especially if it is an entrance exam problem, should have specified TWO points that you are finding the voltage between, and at what point you are starting at.

Now, if only one point is specified in the problem and you are to find the voltage there, the other point is taken to be infinity. This is known as an absolute potential. However, you still need to know if you are starting at infinity and moving a test charge to the point, or if you are starting at the point and moving the test charge to infinity. The "convention", which is not clearly explained, is that when finding absolute potentials you start at the point and move the positive test charge to infinity.

However, absolute potentials are a little bit ugly when dealing with the theoretical construct of the infinite plane, since the E-field due to an infinite plane of charge is constant everywhere in space, which would yield an absolute potential of -infinity.

EDIT: I am an idiot. It says V=0 at the infinte plane!!

Last edited: Aug 25, 2006