# Two positively charged plates - can the electric field be negative inside?

Suppose I have two parallel plates, positively and uniformly charged, a distance $d$ apart and infinite. The surface charge density of the top plate is $\sigma_1$ and the surface charge density of the bottom plate is $\sigma_2$. Then the electric field inside of the plates can be given by (I used Gauss' Law to show this easily): $$\vec{E} = \vec{E_{\text{top}}} + \vec{E_{\text{bottom}}} = - \frac{\sigma_1}{2\varepsilon_0}\hat{z} + \frac{\sigma_2}{2\varepsilon_0}\hat{z} = \frac{\sigma_2 - \sigma_1}{2\varepsilon_0}\hat{z}$$ The issue I'm having here is that because these two plates are both positively charged, I'm sure they need to repel. However, if $\sigma_1 > \sigma_2$, the electric field inside the plates points downwards, which suggests that the upper plate moves towards the lower plate? Isn't that suggesting they're attractive.

As you can see, I'm having a bit of difficulty understanding what this result about the electric field is telling us for how the plates would move in the different cases that arise for $\sigma_1$ and $\sigma_2$, so any help on that would be appreciated.

The electric field from each plate does not act on itself. It only acts on the other plate, and any other charges. The field $E_1$ acts on $q_2$ and the field $E_2$ acts on $q_1$.
If $\sigma_1=\sigma_2$ then the total electric field acting on some other charge is zero, but the forces $q_1E_2$ and $q_2E_1$ acting on each plate are non-zero so the plates still repel.
Whatever you say about $\sigma_1$ and $\sigma_2$, you cannot say they are attracting. If $\sigma_1$>$\sigma_2$, the electric field on the top of top plate is along upper direction and below the top plate the electric field is along downward direction. So the direction of electric field is different in the two sides of the top plate. So we cannot say that they are attracting.