# Electric field between oppositely charged metal plates

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.

• 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 Sep 12, 2022 at 6:40