Capacitor with a dielectric with only one plate / with one plate moving I was wondering how a capacitor like the one in the picture below works. The length of the plates is L and their width is b (parallel to the z axis). The distance between the plates is d. The whole space between the plates is filled with a dielectric of relative permittivity ε_r. At the beginning the capacitor has charge Q. (picture a) 
The lower plate is fixed to the ground but the upper one can be moved. Suppose it has moved a distance x along the x axis (picture b). 
I was wondering what happens to the charge in this situation (assuming it is conserved). What are the charges Q1, Q2, Q3, Q4? What is the energy of this capacitor? I want to use the equation for the energy (derivating it with respect to x) to find the force stopping the movement of the upper plate (In my original problem, the capacitor is situated on a plane and the upper plate slides down due to the force of gravity).   
The part with charge Q3 doesn't have any input, does it? Should I somehow include the part with charge Q4, even though there is only one plate there? I know there is a charge induced in the dielectric here and I can find it knowing Q4 but I don't know how I should treat this part of the capacitor calculating the energy.
EDIT: We assume that d << L and d << b. Also: x >> d but x < L (in fact it will be x < L/2 in the original problem so that the upper plate won't rotate on that plane). For space outside the dielectric ε_r = 1.

 A: To solve this electrostatic problem exactly for an arbitrary plate distance $d$ and relative permittivity $\epsilon_r$ would involve finding the electrostatic field in and around the capacitor with the moving plate which is rather difficult problem. Therefore I assume that the plate distance (and permittivity) are so small that the electric field is predominantly concentrated in the overlapping part $L-x$ of the capacitor.
Under this assumptions, the total initial charge $Q$ (which stays constant during the movement) is always located under the overlapping part of the upper plate because all the electric field is concentrated there. Thus the approximation $Q1=Q$ and $Q2=-Q$ holds for all $x$ until the upper plate is completely removed at $x=L$. This means that $Q4$ and $Q3$ are always zero.
Then the potential electrostatic energy of the capacitor is $$E=\frac{Q^2}{2 C}=\frac{Q^2 d}{2\epsilon_0\epsilon_r (L-x) b}$$ which results in a force  $$F=-\frac{dE}{dx}$$ you have to work against when moving the plate by increasing x. Obviously, for $x$ approaching L this not a good approximation anymore.
A: Consider a simple 2D model, where there are some number of discrete charges distributed uniformly along each plate (=line). Then, to first approximation, each charge experiences a force from all the remaining charges (attractive for the opposite plates, and repulsive for the adjacent plates). These forces add together and cause the charge to move, constrained to the geometry of the plate. The force between the opposite plates is reduced by a factor of $\epsilon_r$ when there is a dielectric between them, or not reduced when there is no dielectric between them, or reduced by a proportion (see below).
A few model assumptions: $x=0.3$, $L=1$, $d=0.05$, $\epsilon_r=3$ and assume $n=30$ charges on the plate. Then, the charges remain broadly distributed on the plate:

This is very unlike the fringe field of a parallel plate capacitor. In that situation, there is nowhere for the charge to go, and it goes broadly uniform because it is constrained to remain on the plates (if that constraint went away, the charges would disperse very rapidly, aka dielectric breakdown). 
In this situation, you are giving the charges a conductor on which to move, and so, of course, they will happily. Yes, they are attracted to the charges on the opposite plate, but this is overwhelmed by the repulsion on the plate. 
Here's what it looks like when $x=0$:

Note that there is a little bit of bunching towards the ends. In this condition, the forces on the charges are balanced, except for the charges on the end, which are experiencing a force trying to get them to shoot sideways off the plate (and of course, there is an attractive force trying to get the charges to the opposite plate).
Even for $x=0.9$ this distributive force dominates:

This is especially so, since the relative permittivity actually reduces the force experienced by the charges on either side of it. The qualitative behaviour does not change much with the other parameters.
A simple way to look at this is as a number of small capacitors in parallel (see the red lines in the diagram):

Each capacitor has a capacitance $\epsilon_r\frac{bdx}{4\pi\sqrt{d^2+x^2}}$ (you can see this methodology on page 159 of this paper) (I'm assuming the dielectric extends, though this can be tweaked). Integrating this across the capacitor gives a capacitance of $C=\epsilon_r\frac{bL}{4\pi\sqrt{d^2+x^2}}=\epsilon_r\frac{A}{4\pi\sqrt{d^2+x^2}}$. This is a change in capacitance by a factor of $\frac{d}{\sqrt{d^2+x^2}}$.
