# Electrical force on parallel plates

In lecture, my professor showed us something peculiar.

2 parallel plates with equal charges lie close on top of each other.

although the electric field in between the plates cancel each other out, the Force on one plate from the other is

$$F = \frac {\sigma^2}_{2 \epsilon}$$

due to

$$\vec{\nabla} \cdot \vec{E} = 0$$
and
$$\rho = 0$$

why is the electric field $$\frac{\sigma}_{2 \epsilon}$$
when the electric field between the plates is 0???

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when the electric field between the plates is 0???
Each plate has the same charge stored on it, so what happens to this charge when the plates are put in contact? Think about a single and isolated charged conductor and how charge distributes itself.

Dear Orthovector:

I would like to give some non-mathematical explanation of the confusion. The two plates carry the kind of charge, and no doubt they will repel each other, and the force is as you put it.

Moreover, because the two plates are exactly symmetric to each other, electrically and geometrically, and properly arranged. Hence, at any point between the two plates the fields generated via the plates are of identical magnitude and opposite direction. Hence the resultant of the field is vanishing.---This doesn't mean that the plates donot repel each other, it indeed means, if we put a test charge at any point between the plates, the resultant of static electric forces is vanishing. The vanishing field is mechanically meaningful to external charge, rather than the plates.

And as to the response and question from Buffordboy23 above, I would also like to make some comments.

Each plate has the same charges stored on it, and if the plates are put in contact, the charges will of course be re-arranged. If isolated without further influence, the charges will gather on the surface of the two plates, the sharper the surface (or the larger the curvature), the denser the charges. The plates is a equipotential body, with the surface equipotential too.

Yours,
Orthotensor and Paravector !