# I Electric field due to plane of charge

1. Apr 27, 2017

### kelvin56484984

Why the gaussian surface is "A" instead of "2A" on the right-hand side?(the photo attached below)
Due to the thickness of the sheet?

I usually find that E=σ/2ϵ0 is being frequently used but we hardly use E=σ/ϵ0.
When can I directly apply the equation E=σ/ϵ0 ?

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• ###### flux.png
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2. Apr 27, 2017

### DuckAmuck

The one on the right is for a conducting surface. There's no charge on the other side of the surface, so there's no factor of 1/2.

3. Apr 27, 2017

### plasmon_shmasmon

I initially suspected that the RHS of the figure was just considering the flux through part of the surface, but then I saw they used the integral notation for a closed surface (meaning you have to consider both sides of the pill-box). Perhaps this shape is meant represent some charged, hollow conductor--therefore there should be no electric field inside and you only have flux through one face of the pill-box.

4. Apr 27, 2017

### Chandra Prayaga

It need not be a hollow conductor. The electrostatic field within the body of a conductor is zero. So there is no field and no flux through the face within the conducting material.

5. Apr 27, 2017

### plasmon_shmasmon

Yes, I agree. But this assumes that the pill-box terminates within the conductor itself. It could be possible that given illustration represents a conducting sheet, and the pill-box has ends on either side. Then I believe the equation would hold only if the sheet were a closed surface (I think that reasoning is correct). I think the illustration is a little unclear.

6. Apr 27, 2017

### Chandra Prayaga

Certainly, the illustration does not clearly show where the pill box ends, but that does not change the result. The pill box is your choice. In principle, Gauss's law is valid for any closed (imaginary) surface, and the pill box is one such surface. By a judicious choice of this Gaussian surface, you can easily calculate the electric field, and in this case, it is advantageous to take one end of the pill box inside the material, because you already know that the electric field is zero there. The shape of the conductor itself does not matter. Whether it is a sheet or not, the pill box argument will be correct, by taking a suitably small cross section area.
I am not sure what you mean by the sheet being a closed surface. If you have a conducting sheet, you can still take one end of the pill box inside the conductor, and the result is still the same.