Charged particle between two parallel likely charged plates, is it affected by the plates? Imagine two parallel conductive plates. Charge up both to have the same amount of positive charge. Then put positive test particle between the two.
The Coulomb's law is an inverse square law, so one might think the positive test particle is repelled from the nearby plane and accelerated towards the middle between the plates making it do an oscillating motion (until it radiates away it's energy due to the acceleration and stops in the middle).
On the other hand, since the two plates have the same charge, there is no voltage between the plates that would do work on the electric charge. So it won't move at all.
I'm a confused here. Is Coulomb's law just a special case for 2 point charges? 
 A: 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.
A: Gauss theorem is of great importance. Those situations, in which the calculation of electric field by applying Coulomb's law or the principle of superposition of electric fields becomes very difficult, the results can be obtained by applying Gauss's theorem with great ease.  
You can notice that electric field due to a point charge decreases inversely as the square of the distance from it (can be calculated from Coulombs law), in contrast, the field due to a line charge falls off as $\frac{1}{r}$, and we find that the magnitude of electric field at a point due to an infinite plane sheet of charge is independent of its distance from the sheet of charge as noticed by hlouis.  
In your case, the charge is going to stay where it was and it mean that potential at every point due to both the charged plates is same, in contrast to the potential due to point charge which depends on the value of r. Coulombs law is one of the special case which can be derived from Gauss's theorem.
A: Yes, Coulomb’s law is only for point charges separated by a distance $r$. The inverse square law is there because of the diverging or converging nature of electric field from a point charge which is the crucial point. Here is how I explain: 
Imagine spherical surfaces with increasing radius around a point charge.  As we increase the radius, the intensity of electric field per unit area of the spherical surface goes on decreasing as $\frac{1}{r^2}$. This means the intensity at any point $r$ decreases as we increase $r$. This is because of the diverging or converging of the electric field of the point charge. An example to understand better: consider particles coming out of a point and they come out with finite numbers forming a spherical surface. As they come out, sphere becomes larger and larger but particles are finite on the surface. So, particle density on the surface of the sphere goes on decreasing. So,the number of particles at $r$ decreases as $r$ increases.
This is what Gauss law tells, when we integrate electric field intensity over the whole surface of larger radius, we should get the same value when we integrate electric field intensity over surface for smaller radius which is $\frac{q}{\epsilon_0}$ which is finite.
But, electric field from charged planes doesn’t diverge or converge in ideal case, it is always perpendicular to the plane since the plane is an equipotential surface. Since electric field doesn’t diverge or converge but are perpendicular to the plane, the electric field remains the same at any point near or far from the plane,  which is $\frac{\sigma}{2\epsilon_0}$ from Gauss law. So, the electric field between two equally charged planes with $+Q$ remains always $0$ , $\frac{\sigma}{2\epsilon_0}-\frac{\sigma}{2\epsilon_0}=0=\frac{dV}{dr}$ or the potential at any point between them remains constant, the test charge would remain wherever it is placed!   
A: It depends on how big your plates are, in the case of two infinite plates the charge wouldn't move. If the plates are finite, the electric field would 'leak' away and the charge would move to the middle and away. In that case there would be a voltage difference, because the voltage near the plates would be positive and the voltage some distance away would be zero so $\Delta V=V_{plate}-V_0=V_{plate}$
