Finding E and V involving 2 planes.

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In summary: Yes, in the text you provide there is a section on setting up the boundaries. In summary, Homework Statement states that two planes have charge density that are not equal. The Attempt at a Solution attempts to solve this problem by using Gauss Law. Gauss Law states that the charge density in each region is the product of the charge density on the two plates and the distance between the plates. The values of the field E1 in each region and of E2 in each region (6 numbers altogether) are calculated, and the fields are added vectorially in each of the three regions. The figure attached has arrows that are meant to represent the fields contributed by each plate. There are only two plates,
  • #36
Great. Do I leave them as distinct quantities, or can I combine them? The questions says "Determine V for all points" so I feel like there should be another form of the answer.
 
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  • #37
The way to present the solution is by listing the potentials in each region, such as

V(x,y,z) = ... for x ≥ 6 cm (top region)
V(x,y,z) = ... for 6 cm ≥ x ≥ -4 cm (in-between region)
V(x,y,z) = ... for x ≤ -4 cm (bottom region)
 
  • #38
Ok I see what you are saying. I think you meant to say z instead of x for all those bounds you gave, but I agree otherwise. And all of those units should be 'V' I believe as well.

Finally for the drawing of the equipotentials, they are just going to be planes parallel to the z planes right?
 
  • #39
Yeah, they're just planes with normal vector in the z direction. I'm confused about the middle region...why wouldn't the "-4" be involved at all in the bounds. I have the same thing for the other two regions and understand why, but I don't understand why the middle region wouldn't be limited by the physical boundary of -4.
 
  • #40
xxbigelxx said:
Ok I see what you are saying. I think you meant to say z instead of x for all those bounds you gave, but I agree otherwise. And all of those units should be 'V' I believe as well.
Sorry, yes I meant to say z instead of x for all the bounds.
 
  • #41
I can't answer for sure, but I would guess it's just because you are only interested in the plate where the E is originating (in this case at z= 6cm)
 
  • #42
noface said:
Yeah, they're just planes with normal vector in the z direction. I'm confused about the middle region...why wouldn't the "-4" be involved at all in the bounds. I have the same thing for the other two regions and understand why, but I don't understand why the middle region wouldn't be limited by the physical boundary of -4.

Because -4 isn't the reference point. If you wanted the capacitance then you would need the potential difference between V(6) and V(-4), otherwise it doesn't matter. Don't worry, the boundaries solve any issues of ambiguity.

Lastly, don't worry, this is a textbook problem. It's meant for you to learn about EM. In reality, with no approximations, you would always set the reference voltage at infinity. Griffiths has a good discussion of this.
 

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