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

  • Thread starter Thread starter mmfiizik
  • Start date Start date
mmfiizik
Messages
7
Reaction score
0
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
1762851673578.webp


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:

1762853476941.webp

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:
1762851711652.webp
 
Physics news on Phys.org
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.
 
Thread 'Chain falling out of a horizontal tube onto a table'
My attempt: Initial total M.E = PE of hanging part + PE of part of chain in the tube. I've considered the table as to be at zero of PE. PE of hanging part = ##\frac{1}{2} \frac{m}{l}gh^{2}##. PE of part in the tube = ##\frac{m}{l}(l - h)gh##. Final ME = ##\frac{1}{2}\frac{m}{l}gh^{2}## + ##\frac{1}{2}\frac{m}{l}hv^{2}##. Since Initial ME = Final ME. Therefore, ##\frac{1}{2}\frac{m}{l}hv^{2}## = ##\frac{m}{l}(l-h)gh##. Solving this gives: ## v = \sqrt{2g(l-h)}##. But the answer in the book...