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Two large metal plates face each other and carry charges with surface charge density $+\sigma$ and $-\sigma$, respectively, on their inner surfaces. Find $\mathbf{E}$ between them.

The answer is $$E = \frac{\sigma}{\varepsilon_0}$$ but with superposition it seems it should be twice that: $$E = \frac{2\sigma}{\varepsilon_0}$$ ($\sigma$ is the area charge density on the plates, and $\varepsilon_0$ is permittivity constant).

I'm self-studying from Physics, 4th Edition, by Halliday, Resnick, Krane. They are very explicit about the difference between conducting and non-conducting infinite sheet's electric field.

For non-conducting infinite sheet they give $$E = \frac{\sigma}{2\varepsilon_0}$$

For conducting infinite sheet they give $$E = \frac{\sigma}{\varepsilon_0}$$ because the other side of the metal sheet has equal charge, so you have superposition of the two sides.

In the space between the oppositely charged plates, each conducting metal plate creates a field of $$E = \frac{\sigma}{\varepsilon_0}$$ both pointing in the same direction (from positive towards negative). Which sums via superposition to $$E = \frac{2\sigma}{\varepsilon_0}$$

If these were non-conducting plates, then each plate contributes $$E = \frac{\sigma}{2\varepsilon_0}$$ which sums via superposition to $$E = \frac{\sigma}{\varepsilon_0}$$

But this problem specifies "two large metal plates", not non-conducting plates.

opposite charged plates

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  • $\begingroup$ Find electric field of a single infinte charged sheet in an otherwise empty space, then add another sheet, you'll see the error in your above analysis $\endgroup$ Sep 12, 2022 at 6:40

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It is because Gauss's law gives the net electric field and not due to a single component . If you draw a gaussian surface to calculate the electric field between the plates, then what you get is net electric field , irrespective of your choice of gaussian surface , instead of just the electric field due to a single plate .

It is because , if there was only one plate and the other plate being removed , the surface charge density is halved , as other charges would go to the outer face of conductor in absence of other plate.

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  • $\begingroup$ So the outside surfaces have zero charge in this case because all the charge is attracted to the inner faces. Makes sense. Now I'm wondering how to calculate when the charges are unequal, like one sheet has +2q and the other has -q. But that's a different problem. $\endgroup$
    – owler
    Sep 12, 2022 at 15:52
  • $\begingroup$ @owler Yes , that's a different problem. But you can do it simply by assigning different variables of charges present on each surface and then use the fact that inside a conductor, the resultant field due to all charges will always be zero . This will help you to make equations with your variables and you'll get the charge distribution . Hope it helps . $\endgroup$
    – Abbas
    Sep 12, 2022 at 16:56
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    $\begingroup$ yes, very helpful. With your ideas I was able to work out the problem where one sheet has a charge of +k q and the other has -q, and it all makes sense now. $\endgroup$
    – owler
    Sep 12, 2022 at 18:15
  • $\begingroup$ @owler Welcome . I am glad to help :) $\endgroup$
    – Abbas
    Sep 13, 2022 at 2:56

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