Two-layer dielectric, four-capacitor model vs three-capacitor model

[mmfiizik]

Homework Statement

The lower dielectric is moved out of the arrangement by x in the +ex direction, as shown in
the right-hand image. The value x changes in the value range 0 ≤ x ≤ a. Calculate the total capacitance of the arrangement as a function of x. The boundary effects should be neglected for the calculation. The
surrounding air is assumed to be er,L = 1.

Relevant Equations

Parallel plate capacitor

I am working on part (f) of a exercise. The setup is a square parallel-plate capacitor with plate side a and separation d. Initially the gap is filled by two layers on top of each other. The upper half has relative permittivity epsilon_{r1} and the lower half has epsilon_{r2}. Then the lower dielectric is moved in the +e_x direction by a distance x. As a result, on the left region of width x, epsilon_{r1} stack becomes over air, while on the right region of width a-x the stack remains epsilon_{r1} over epsilon_{r2}.

My interpretation was to split the upper plate area into two parallel stripes that share the same plate voltage. The left stripe has area A1=ax and is a series combination of epsilon_{r1} and of air (epsilon_{rL} = 1). The right stripe has area A2=a(a−x) and is a series combination of epsilon_{r1} and of epsilon_{r2}. And those two in series are then connected in parallel. However when I calculate capacitance this way, I get different answer.

The official solution instead appears to treat the entire top slab epsilon_{r1} as a single capacitor Cr1 placed in series with (Cr2∥CL), where CrL represents the air part. I struggle to see why the whole top slab can be replaced by one series element when its lower boundary sits partly over air and partly over epsilon_{r2}. In my view the left and right stripes are are regions in parallel (because of same voltage drop?). So I am confused in how should I calculate capacitance. Even if I am not restricted in splitting the material and I end up with configuration like this:

how do I know if it is (C1 || C2) + (C3 || C4) or (C1 + C3) || (C2 + C4) ? By "+" I mean series.

Thanks for any insight.

Solution:

 

[mjc123]

I think you are right. The book solution implicitly assumes that the voltage at the bottom of slab 1 is constant across the width, which is not the case. You have to treat it as two series combinations in parallel.

 

[mmfiizik]

I think you are right. The book solution implicitly assumes that the voltage at the bottom of slab 1 is constant across the width, which is not the case. You have to treat it as two series combinations in parallel.

But since the top slab doesn’t move, top plate's surface-charge density (or charge distribution) shouldn’t change. Then the normal displacement at the plate is the same everywhere (D=σ=const). Because the top layer is the same material ε_r1 across the whole width, the electric field there would also be the same, E1=D/(ε_0*ε_r1), so the drop across the top layer ΔV1=(d/2)E1 should be identical at every lateral position. Wouldn’t that make the dielectric interface with ε_r1, ε_r2 and ε_rL (air) an equipotential surface?

 

[mjc123]

But I don't accept your premise. You assert that the top plate's charge distribution doesn't change, therefore the electric field doesn't change. Why? I say that the electric field changes, depending on what's between the plates, and therefore the charge density will change too. The key point is that to have the same charge density on the top and bottom plates (at corresponding x) the field must satisfy
epsilon_tE_t= epsilon_bE_b (It's still not letting me insert symbols!)
as well as E_td/2 + E_bd/2 = deltaV
Therefore E_t, and the potential at d/2, will vary depending on whether the bottom is filled with slab 2 or air at that point.

 

[Steve4Physics]

But since the top slab doesn’t move, top plate's surface-charge density (or charge distribution) shouldn’t change. Then the normal displacement at the plate is the same everywhere (D=σ=const). Because the top layer is the same material ε_r1 across the whole width, the electric field there would also be the same, E1=D/(ε_0*ε_r1), so the drop across the top layer ΔV1=(d/2)E1 should be identical at every lateral position.

I don’t think that’s correct – there will be discontinuities in charge-densities and field-strengths as you move left to right. To add what @mjc123 said, I think the system should be visualised like this:

 

[mmfiizik]

Thank you all for explanations. I tried to prove this to myself by contradiction, but it might be an oversimplification.

If we assume that there is uniform charge density on both plates, it would mean that at every point out of that plate displacement field $\text{D}$ would be the same. Then voltage drop across capacitor left part would be:
$$\Delta V=U_{0}=\frac{Dd}{2\varepsilon_{0}\varepsilon_{r,1}}+\frac{Dd}{2\varepsilon_{0}\varepsilon_{r,L}}$$
And for the right part:
$$\Delta V=U_{0}=\frac{Dd}{2\varepsilon_{0}\varepsilon_{r,1}}+\frac{Dd}{2\varepsilon_{0}\varepsilon_{r,2}}$$
But since $U_{0}$ is fixed by the voltage source, and $\varepsilon_{r,L}\neq \varepsilon_{r,2}$ (they were equal in the first part of the task), $\text{D}$ cannot be the same for right and left capacitor sides. Then charges on the metal plate have to redistribute so that $D_{right}\neq D_{left}$. But then voltage drop in the top slab is not the same anymore:
$$\Delta {V_{1,right}}=\frac{D_{right}d}{2\varepsilon_{0}\varepsilon_{r,1}}$$
$$\Delta {V_{1,left}}=\frac{D_{left}d}{2\varepsilon_{0}\varepsilon_{r,1}}$$

Therefore the internal interface is not equipotential. So my premise was indeed incorrect and we have two series stacks in parallel.