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FrogPad
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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?
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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