EM Troubles - Conducting body / potential

In summary, the conversation is discussing a project involving a conducting plate in free space and determining the potential with respect to a reference point using an integral expression. The participants also discuss the challenges of solving the problem due to the varying charge distribution on the surface of the conductor. The project will be solved numerically using the Method of Moments, and the participants are still working on understanding and solving the problem.
  • #1
FrogPad
810
0
So I have an upcoming project that I started working on. I thought I had the foundation to actually tackle the task at hand, but now that I've started it, I realized I'm kind of weak with some fundementals. If someone would be so kind to read what I am going to type, and add to it, or tell me what I'm doing wrong, I would be very appreciative.

So...

Say we have a conducting plate in free space. If we set a reference point far enough away and measure the potential with respect to this point we have,

[tex] V(\vec R) = \int_V \frac{\rho_V(\vec R')}{4 \pi \epsilon_0 R} \, dv' [/tex]

Where:
[tex] \vec R [/tex] is the distance from the orgin to a point of interest (where we will measure the potential)
[tex] \rho_V [/tex] is the volume charge density of the plate.
[tex] \vec R' [/tex] is the distance from the orgin to a charge (which are located on the plate.
[tex] R = |\vec R' - \vec R| [/tex]
[tex] dv' [/tex] is the differential volume.

Now since we have a conducting plate, the voltage (is this part right?) on the plate (with respect to our reference point) is constant.

If we call this constant voltage [itex] V_0 [/itex] then we have,

[tex] V(\vec R) = V_0 = \int_V \frac{\rho_v (\vec R')}{4 \pi \epsilon_0 R} dv' [/tex]
If (could someone help me with this set notation here too?),
[tex] \vec R \in S [/tex]
Where: [itex] S [/itex] represents the surface of the plate.

(I'm trying to say, [itex] \vec R [/itex] points to a point that is on the surface of the plate.

Thus, as long as [itex] \vec R [/itex] points to a location on the surface, the integral will evaluate to a constant, [itex] V_0 [/itex].

Is my understanding correct?
 
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  • #2
conductors are supposed to be equipotential, so i think yes. that expression is correct
 
  • #3
Regarding your question, you should use [tex] \sigma dA [/tex] rather than [tex] \rho_v dv[/tex]. If your plate is a conductor, then all the charges are at the surface in an electrostatic case.

Also, you might want to consider solving the Laplace equation to find V. In many cases, this approach is much easier (and practical, as you usually have control over the values of V at the boundaries).
 
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  • #4
FrogPad said:
So I have an upcoming project that I started working on. I thought I had the foundation to actually tackle the task at hand, but now that I've started it, I realized I'm kind of weak with some fundementals. If someone would be so kind to read what I am going to type, and add to it, or tell me what I'm doing wrong, I would be very appreciative.

So...

Say we have a conducting plate in free space. If we set a reference point far enough away and measure the potential with respect to this point we have,

[tex] V(\vec R) = \int_V \frac{\rho_V(\vec R')}{4 \pi \epsilon_0 R} \, dv' [/tex]


Is my understanding correct?
No.
You can't simply use that Coulomb's law integral to find V(r) in the presence of any conductor, because you can't determine rho until you have solved the problem. That is because the charge positions itself on the surface of any conductor to make the conductor an equipontential.
You will have to use some Greens function method. If the conducting plane can be considered infinite, you could use images.
 
  • #5
Meir Achuz said:
No.
You can't simply use that Coulomb's law integral to find V(r) in the presence of any conductor, because you can't determine rho until you have solved the problem. That is because the charge positions itself on the surface of any conductor to make the conductor an equipontential.
You will have to use some Greens function method. If the conducting plane can be considered infinite, you could use images.

This problem will actually be solved numerically.

The conducting plane is finite, and it is supposed to be solved with the Method of Moments. I believe that method uses Green's function (something I have not studied). The professor gave us an example of the setup (using MoM) for a 1-dimensional case, and we are expected to come up with something similar for a 3-dimensional case.

My understanding is limited at this point, as the project is not due for another month. However what I've read thus far, is yes [itex] \rho_S [/itex] or [itex] \sigma [/itex] or whatever you call it is NOT constant, and arranges itself (as you said) so that the voltage on the plate is constant.

I wanted to make sure that integral expression I wrote is correct.
So it is correct right? It's just we don't know what the charge function is at this point?

So now if we restrict the domain to points on the conductor, the voltage will be constant. This leaves the conduction function to be solved.

This is the part I am working on right now, and need to figure out. I wanted to make sure I was cool, up to this point.

The little I know this far is,

[itex] \rho_S [/itex] (I'm comfortable with this notation, I hope this is ok) is broken into a piecewise function, then we can assume [itex] \rho_{Si} [/itex] is constant on each "square" of the conducting plate. A condition must be set that says that [itex] \rho_S = 0 [/itex] if not on the plane.

I still have more to do for sure...
is it cool if I use this thread with more questions regarding this project?

thanks everyone!
 

1. What is a conducting body?

A conducting body is a material that allows electricity to flow through it easily. These materials typically have a high concentration of free electrons which can move easily in response to an electric field. Examples of conducting bodies include metals, graphite, and some solutions containing ions.

2. How does a conducting body differ from an insulator?

A conducting body is essentially the opposite of an insulator. While a conducting body allows electricity to flow through it, an insulator does not. Insulators have a low concentration of free electrons and therefore do not allow for the easy movement of electricity.

3. What is potential in relation to conducting bodies?

Electric potential, also known as voltage, is a measure of the difference in electric potential energy per unit charge between two points. In the context of conducting bodies, potential refers to the difference in electric potential between the body and its surroundings. This difference in potential is what drives the flow of electricity through the conducting body.

4. How does the shape and size of a conducting body affect its potential?

The shape and size of a conducting body can affect its potential in several ways. Firstly, a larger conducting body will typically have a larger surface area, allowing for more charge to be stored. Additionally, the shape of the conducting body can affect the distribution of charge, which in turn can affect the potential. For example, a sphere will distribute charge evenly on its surface, while a cylinder will have a higher concentration of charge at its ends.

5. What are some common EM troubles associated with conducting bodies?

One common EM trouble associated with conducting bodies is the buildup of static electricity, which can cause electrical shocks and damage to electronic devices. Another issue is the interference of electromagnetic fields with electronic devices, which can cause malfunctions or disruptions in their performance. Additionally, conducting bodies can also create EM shielding, blocking or redirecting electromagnetic waves and causing communication issues for devices such as cell phones or radios.

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