Yes the additional positive charge to one plate increases electric field behind the other one.
To see this let's look at general Gauss's law:
$$\oint_{S}\vec{E} \,d\vec{A}=Q/\epsilon_0$$
The $S$ is the Gaussian surface which we are free to choose therefore it could include both plates even if the surface goes to infinity with both plates.
The charge Q however could be written as charge density $\sigma$ times surface area or in general:
$$Q = \int_A \sigma \, dA $$
where $\sigma$ could be $\sigma_1$, $\sigma_2$ or $\sigma_1 + \sigma_2$.
Now we are ready to use symmetry of configuration. Lets assume that field is homogeneous at the middle of of both infinities and rewrite the left side of Gauss law as:
$$\oint_{S} \vec{E}\, d\vec{A} = \oint_{S_{in}} \vec{E}\, d\vec{A} + \oint_{S_{out}} \vec{E}\, d\vec{A} $$
The right side with the same division:
$$\int_A = \int_{in} +\int_{out}$$
Now we are ready for the main part:
$$\oint_{S_{in}} \vec{E}\, d\vec{A} + \oint_{S_{out}} \vec{E}\, d\vec{A}
=\int_{in}\sigma \, dA + \int_{out}\sigma \, dA$$
And the Gauss law says that charges which is outside from closed surface does not affect the flux or $\vec{E} = \vec{E}_{in} + \vec{E}_{out}$ and the $\oint_{out} \vec{E}_{in} \, dA = 0 $ which makes to satisfy:
$$\oint_{S_{out}} \vec{E}\, d\vec{A}
= \int_{out}\sigma \, dA$$
And the only thing which is left is always satisfied:
$$\oint_{S_{in}} \vec{E}\, d\vec{A}
= \int_{in}\sigma \, dA$$
which for homogeneous field means:
$E_{top} + E_{bottom} = \sigma / \epsilon_0$
where the signs for field is positive if it is directed out of surface.
If the rods were really far apart then the positive charge would be equally distributed throughout each rod. If you push the rods together then the new equilibrium involves fewer charges bunched around the closer points and more of them at the far ends; the energy you exert is the energy it takes to move these charges around. Even if we didn't consider the atomic nature of "contact" and pretended we had a continuous material, you would just find that the point of contact had zero positive charge on it, with the positive charge instead being distributed more toward the far ends.
Presumably if the charges were literally totally immobile and these were literally continuous and smooth materials then you couldn't make them touch for the reasons you stated. But in reality:
- the charges all have some ability to move (even if it's just on atomic distances)
- They don't actually come within zero distance from each other
- The surface is bumpy on a microscopic level so there are probably extremely few actual contact points
- You're unlikely to actually be touching two charged atoms together given the above and the sheer number of atoms in the material
- Even if you did manage to precisely line them up to do 4, all you'd have to do is overcome the force it would take for the atomic lattice to rearrange and move the positive charges out of the way.
Also the force on two electrons a nanometer apart is given by $\frac{ke^2}{(10^{-9})^2}=2.309\cdot 10^{-10} N$. So even if you did manage to circumvent all of the above, you'd probably only have a few contact points at most each contributing a fraction of a nanonewton. Due to the incredibly small charge of the electron it would take a lot more than that to generate a significant force.
Best Answer
Coulomb's law is indeed a special case between two point charges. to find the force between a point charge and a plate, you would have to integrate the equation over the plate surface to calculate the contributions from all infinitesimal charge elements.
It's more practical to figure out what the electric field is, and then use $\vec{F}_e = q\vec{E}$ to find if there is a force applied to your test charge.
The electric field from an infinite plate with uniform surface charge density is given by $\vec{E} = \frac{\sigma}{2\epsilon_0}\vec{a}_n$ where the $\vec{a}_n$ vector is pointing away from the plate. Therefore, if you have two parallel plates (sufficiently large compared to the distance between them to be considered infinite) with the same $\sigma$ charge density, the electric field between the two will be null, and no force will be exerted on the test charge.