Three parallel plates -- find the electric field

  • #1
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Homework Statement


Find the field at A.
three-plates-exam-jpg.jpg


Homework Equations


##\oint E\cdot dA = Q_{enclosed}/\epsilon_0##

The Attempt at a Solution


My first intuition was to do a Gaussian cylinder from A to the middle of the bottom plate. My logic is that the field inside the bottom plate is 0, so I'd have (from Gauss' Law) ##E(\pi r^2) = (10 - 2 - 1)\sigma \pi r^2/\epsilon_0##, so ##E = 7\sigma / \epsilon_0##. However, the correct answer is with a coefficient of 1.5, which I get if I do a Gaussian cylinder from A to D.
This also got me thinking: if I do a Gaussian cylinder from middle of the top plate to the middle of the bottom plate, my understanding is that field is 0 at the top and bottom of my Gaussian surface. Yet, ##E(2\pi r^2) = (10 - 1 - 1)\sigma \pi r^2/ \epsilon_0## and this suggests to me that field is nonzero inside the plates. What is wrong with my understanding here? Does this suggest that field is nonzero inside the plates?
 

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Answers and Replies

  • #2
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I am not sure a cylinder is the right geometry for this problem. Maybe you need a cube.

For sure, if the plates are made of metal then the field is always zero inside them.
 
  • #3
kuruman
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The straightforward way to do this is first to use Gauss's Law once to find a general expression for the field due to one plate with surface charge density σ. Then just use superposition for the field in any region. A vector diagram of the direction of the field due to each plate in the region of interest would be extremely helpful.
 
  • #4
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Slight aside: when calculating ##\sigma## here, is this ##Q/A## where ##A## is the area of one side of the sheet? Or does this include both sides together? What I mean by this is: if a plate contained 4 C of charge, and one face of the plate was 2 m^2 in area, would ##sigma## equal 1 or 2 (coulombs/m^2)?
I ask because, here, it seems that sigma is calculated using total charge / area of one face. However, in the last example on this page (https://physics.info/law-gauss/), it seems to be calculated as total charge / area of both faces. Which is correct/the more typical convention?
 
  • #5
kuruman
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The comment by @Gene Naden is a distraction. If you had conducting plates, the figure would be showing different charge densities on the two sides of each plate, so it is a good assumption that the given charge distributions are not free to move. Therefore, if you draw the standard Gaussian pillbox through the plate, the enclosed charge is qencl = σ dA where dA is the area of the face. There are two faces through which electric field lines pass so the electric flux is Φ = 2 E dA.
However, in the last example on this page (https://physics.info/law-gauss/), it seems to be calculated as total charge / area of both faces. Which is correct/the more typical convention?
It's not a matter of convention; it's a matter of what you have here as indicated in the figure. You have case I, "planar symmetry nonconducting plane of infinitesimal thickness with uniform surface charge density σ".
 
  • #6
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The comment by @Gene Naden is a distraction. If you had conducting plates, the figure would be showing different charge densities on the two sides of each plate, so it is a good assumption that the given charge distributions are not free to move. Therefore, if you draw the standard Gaussian pillbox through the plate, the enclosed charge is qencl = σ dA where dA is the area of the face. There are two faces through which electric field lines pass so the electric flux is Φ = 2 E dA.

It's not a matter of convention; it's a matter of what you have here as indicated in the figure. You have case I, "planar symmetry nonconducting plane of infinitesimal thickness with uniform surface charge density σ".
But, for instructive purposes, wouldn't the field of case 2 be half their answer?
I get this with ##E(2\pi r^2) = \sigma \pi r^2 / \epsilon_0## using a gaussian cylinder extending above the thick plate to below it.
 
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  • #7
haruspex
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wouldn't the field of case 2 be half their answer?
Their case 2 is just case 1 doubled up. Each surface of the thick plate is equivalent to the infinitely thin sheet. Since each face has the same charge density as in case 1, any given area of the thick sheet carries twice the charge.
 
  • #8
rude man
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The problem statement is confusing. On which sides are the indicated surface charges? For example, is the surface charge +5σ on the top and another +5σ on the bottom, or is there zero charge on the bottom, or ...?
 
  • #9
kuruman
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On which sides are the indicated surface charges?
The "plates" are not conductors. They are infinitesimally thin distributions of "pasted on" charges. That's why the + and - signs are drawn as they are.
 
  • #10
rude man
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Extra credit to OP: what if the top charge layer had been -2σ instead of -σ, the others the same as before?
 

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