Gauss' Law for a conducting / non-conducting sheet

  • #1
laser
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Homework Statement
Is E the same
Relevant Equations
Gauss' Law
c0b995e3-e333-4a70-9174-99c68f98eae2.jpg

The first image is for a conducting sheet (part of it anyway), the second is for a nonconducting sheet. Gauss' law seems to tell me that the electric field strength are different - they differ by a factor of two. Is this true?

The charge enclosed in both of them are the same, and my intuition tells me that the electric field strengths should be the same. Would appreciate some straightforward and to the point answers, as have been scratching my head on this one for a while.
 
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  • #2
Suppose you have a large, isolated conducting sheet. Suppose you could arrange for a uniform charge distribution to be initially placed on the upper surface of the sheet as shown.
1711836369348.png

Can the isolated conducting sheet be in electrostatic equilibrium in this condition? How will the charge redistribute to bring the conductor into electrostatic equilibrium?
 
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  • #3
TSny said:
Suppose you have a large, isolated conducting sheet. Suppose you could arrange for a uniform charge distribution to be initially placed on the upper surface of the sheet as shown.
View attachment 342574
Can the isolated conducting sheet be in electrostatic equilibrium in this condition? How will the charge redistribute to bring the conductor into electrostatic equilibrium?
Thanks for making this clear. So what you are saying is that the charge will redistribute, and then you will have electric field from the top surface and the bottom surface, thus doubling the electric field. Interesting how I couldn't find this mentioned in my notes/book/online.
 
  • #4
laser said:
Thanks for making this clear. So what you are saying is that the charge will redistribute, and then you will have electric field from the top surface and the bottom surface, thus doubling the electric field.
If you want the same charge density on the upper surface of the isolated conducting sheet as you have on the surface of the isolated nonconducting sheet, then you have to put twice as much total charge on the conducting sheet:

1711842691323.png


The electric field at point p for the isolated conductor will then be twice the field at point p' of the isolated nonconductor. The electric field at p of the conductor is the superposition of the fields from the charge on the upper and lower surfaces of the conductor.

laser said:
Interesting how I couldn't find this mentioned in my notes/book/online.
Each situation should be thought through on its own. For example, you could have two conducting sheets with equal and opposite charge, as shown on the left. Now, the bottom sheet is in electrostatic equilibrium without any charge on the bottom surface of this sheet. Likewise, the upper sheet is in equilibrium.

1711843453940.png


The electric field at p for the conductors will be twice the field at p' for the single nonconducting sheet. Check that this is consistent with Gauss's law applied to the pill boxes shown:

?hash=aa9c83e60199c7d189ca53cfbc88c6f2.png


Another situation is the comparison of an isolated, conducting spherical shell with an isolated, nonconducting spherical shell. Each shell has the same uniform charge density on the outer surface. How do the fields compare at p and p'?

1711844499609.png
 

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  • #5
Thanks for providing more examples!
TSny said:
Another situation is the comparison of an isolated, conducting spherical shell with an isolated, nonconducting spherical shell. Each shell has the same uniform charge density on the outer surface. How do the fields compare at p and p'?
I would say that they are the same, as due to Newton's shell theorem, there is no electric field inside a hollow spherical shell.
 
  • #6
There are very different assumptions going in here.

In the one case you have a conductor that is assumed to have some fixed surface charge density. Solely by the fact of being a conductor, the internal field is zero at equilibrium as it will otherwise drive a current that will push the field to zero by rearranging the charges. The rest of the setup must be arranged such that the conductor indeed gets this constant charge density.

In the second case you have an assumed constant charge surface density. You also have the assumption of the field being symmetric under reflections in the surface. While somewhat reasonable, this is not necessarily true. In particular when you have fields that do not vanish at infinity, there are some ambiguities to this. In cases such as the surface charge, you can solve this by requiring the boundary conditions to be symmetric under reflections in the plane as well, but this is not always possible for other charge configurations. The most illustrious example of this being a constant charge density in all of space.

It is perfectly possible to apply an external electric field such that the field below the charge density becomes zero even in the case of a non-conducting surface charge density.

Ultimately, what Gauss' law does tell you is that there is a discontinuity in the component orthogonal to the surface charge density and gives the size of this discontinuity.
 
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