Force between the plates of a capacitor with dielectric slab inserted

[Pushoam]

Like I said, I cannot picture this from a word description.. Please post a picture of the entire Gaussian surface that you have chosen.

Here is the Gaussian surface I was talking of. It includes charge of upper plate and upper part and lower part of dielectric. Electric field due to the dielectric outside the dielectric is 0. Hence, the electric field at bottom plate (which is outside the dielectric) is only due to upper plate, not the dielectric.

The bound charges and the free charges are not mixed with each other. These are divided by a surface (which is invisible here as I cannot draw it here). The bottom part of my Gaussian surface is here.
Now, as this surface is close to the facing surface of bottom plate, the electric field on this surface is close to the electric field acting on the charges of bottom plate.

Just above the bottom part of Gaussian surface, $\vec E = \frac{\vec E_0}K$ and just below the bottom part of Gaussian surface, $\vec E = \vec E_0$. The bottom plate is just below, hence here, $\vec E = \vec E_0$. The bottom part of Gaussian surface is at the point of discontinuity.

This approach shows that the force remains same in both cases (1) and (2). Which mistake am I making in this approach?

 

[TSny]

Here's another way to argue that the force is the same with or without the dielectric. A standard problem in the textbooks is to work out an expression for the capacitance $C$ of a parallel plate capacitor with a dielectric slab that doesn't fill the space between the plates.

I believe the answer is $$\frac 1 C = \frac{1}{\varepsilon_0 A}\left( x + \frac{d}{K}\right)$$ where $A$ is the plate area.

So, when the capacitor is charged with $\pm Q$ on the plates, the energy stored is $$U = \frac{Q^2}{2C} = \frac{Q^2}{2\varepsilon_0A}\left(x + \frac d K \right)$$

Moving the left plate to vary $x$, $U$ will change. The magnitude of the force on the left plate is then $$|F| = \frac {dU}{dx} = \frac{Q^2}{2\varepsilon_0 A}$$ which is independent of $x, d,$ and $K$. So this will be the force on the left plate in the limit of $x \rightarrow 0$ and the force will be independent of $K$.

 

[Pushoam]

I think there is a thin layer of vacuum between the dielectric and the plate. This thin layer doesn't matter while calculating potential energy of the system (capacitor + dielectric) as integral of the force over this small distance (leading to thinness of the layer) is negligible. But, it does matter while calculating force as the electric field in this layer is unaffected by the presence of dielectric.

The expression for force which @TSny has got is calculated in the region without dielectric (here $x\rightarrow 0$ means that the plate goes close to the dielectric, yet it is slightly out of the dielectric, not touching it. So, here d acts as a constant and gets out of differentiation.

Instead of putting the plate at a distance x from dielectric in vacuum and then differentiating its potential energy w.r.t. x, let's try to move the plate inside the dielectric (which is possible when the dielectric is fluid).
Here, $U = \frac{Q^2}{2A\epsilon_0}\frac{x}{K}, F = {-\frac{Q^2}{2KA\epsilon_0}}$

But, according to the question, the dielectric is a slab and hence, the plate cannot move inside the dielectric causing "d" to act as a constant and there is a thin layer of vacuum between the slab and the dielectric, which corresponds to $F = -\frac{Q^2}{2A\epsilon_0}$ satisfied by both approaches (potential energy and Gauss's law).