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I am very confused about plates (conductors/insulators) and applying Gauss law. It seems like gauss law for the isolator is very often derived wrongly.

In short: if you have an infinite plate (insulators) then you can place a cylinder perpendicular to the surface and often the following derivation follows:

Gauss law:

$$ \oint_S EdA = \frac{Q_{\text{enclosed}}}{\epsilon_0} $$

$$ E 2(\pi r^2)= \frac{\sigma (\pi r^2)}{\epsilon_0} $$

$$ E = \frac{\sigma}{2\epsilon_0}$$

However even for a very thin infinite plate you enclose two surfaces, so you then you find for an insulator: $$ E 2(\pi r^2)= \frac{\sigma 2(\pi r^2)}{\epsilon_0} $$ $$ E = \frac{\sigma}{\epsilon_0}$$

The confusing thing is it is usually followed by a derivation with a conductor, giving the last equation.

So my specific question is, is the usual derivation for the insulating plate wrong? The proper derivation you have to use charge density per volume instead of surface density

$$ E 2(\pi r^2)= \frac{\rho d (\pi r^2)}{\epsilon_0} $$ $$ E = \frac{\rho d }{2\epsilon_0}, \rho d = \sigma $$ $$ E = \frac{\sigma}{2\epsilon_0}$$

The first derivation can be done for a conductor I think, because all charge is on the surface.

bananenheld
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  • It is an insulator, not an isolator. What does it mean to enclose two surfaces? What two surfaces? Do you think the charge density of a charged sheet is $\sigma$ per side? Normally the charges along the entire thickness would be included in $\sigma$. Why do you think the conductivity of the sheet is relevant? There also seems to be a factor of 2 missing in your last equation. – Puk Apr 23 '22 at 20:29
  • When finding the field of 2 infinite charged plates, best to find the fields using gauss law individually, and then use the principle of superposition to find the total field – jensen paull Apr 23 '22 at 20:32
  • Possible duplicates: https://physics.stackexchange.com/q/65191/2451 and links therein. – Qmechanic Apr 23 '22 at 20:38
  • You are enclosing two surfaces with surface density $\sigma$, which I would say that you need to multiply $\sigma$2A – bananenheld Apr 23 '22 at 20:46
  • Those links don't answer the question ... because my question is about that, those are the wrong derivations. – bananenheld Apr 23 '22 at 20:49
  • They are derivations of fields that assume charge is moddeled as surface charge. If you want a derivation of the field of a plate with thickness, modeling the entire plate with UNIFORM charge. Then use gauss law, but instead of Q_{enclosed} being \sigma * area, use \rho * volume. Where if your using your cylinder gaussian surface then Q enclosed is $\rho * \pi r^2 * d$ which is what you have done. In either case modelling it with or without thickness, the fields produced by either are exactly the same. The total charge of the plate is the thing that is important. Plug number in. – jensen paull Apr 23 '22 at 21:04
  • as long as the plate has the same charge,Thickness doesn't matter ( for an infinite plate). The plates have have same charge Provided $\sigma = \rho d$ – jensen paull Apr 23 '22 at 21:06
  • Obviously real plates aren't infinite, so both formulas fail. However for capacitors, like you stated, charge resides on the surface, so can easily be modelled as $\sigma$ instead of $\rho$ – jensen paull Apr 23 '22 at 21:08
  • yes but that is exactly my question. When you use $\sigma$ approach like all the derivations you get the wrong result! Because you have to include two surfaces! – bananenheld Apr 24 '22 at 04:43

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