Capacitor with two dielectrics inserted diagonally

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Homework Help Overview

The discussion revolves around calculating the capacitance of a capacitor with square plates, incorporating two different dielectrics inserted diagonally. The participants explore the implications of varying the proportions of the dielectrics and the integration required to derive the capacitance expression.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss various scenarios for the dielectric materials, including cases with 100% of one material, mixed fractions, and the impact of different distributions along the capacitor's length. There are attempts to set up integrals for calculating capacitance based on elemental areas and dielectric distributions.

Discussion Status

Some participants have provided guidance on forming elemental areas and integrating to find total capacitance. There is an ongoing exploration of the mathematical expressions derived, with some participants noting corrections and adjustments to the equations presented.

Contextual Notes

There are mentions of specific variables and constants, such as ε and κ, with suggestions to clarify their definitions. The discussion also touches on the potential implications of equal dielectric constants in the derived equations.

carlyn medona

Homework Statement


A capacitor has square plates of length a separated at a distance d. Dielectrics are inserted as shown in the figure. Find the capacitance

Homework Equations

The Attempt at a Solution

[/B]
I know that I have to integrate, but can't get an expression
 

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What would be the capacitance with 100% of material 1? With 50% of material 1 (and 100% of material 2)? With a fraction x of material 1?

What would be the capacitance if half of the length had x% of material 1 and the other half had y%?

These steps should help to find the right expression.
 
carlyn medona said:

Homework Statement


A capacitor has square plates of length a separated at a distance d. Dielectrics are inserted as shown in the figure. Find the capacitance

Homework Equations

The Attempt at a Solution

[/B]
I know that I have to integrate, but can't get an expression
Take an element dx at a distance x from any end for this caculate dc
 
You can form an elemental area dx dy on the bottom plate, extend the area along the z axis
as a column to the top plate, compute the capacitance of this column as the series connection of two capacitors, then add (integrate) the parallel columns to find the total capacitance.

Actually, it is easier to form dx columns of equal dielectric distribution, then there is only one integral to evaluate.

PS thanks @cnh1995 for the vote!
 
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rude man said:
Actually, it is easier to form dx columns of equal dielectric distribution, then there is only one integral to evaluate.
Yes, that is how I solved it.
 
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Okay I got

dc = εa^2k1k2dx ÷(d (xk1 -xk2+k2a) and after integration c
C = εa^2k1k2ln(k1÷k2)÷(d(k1-k2)
 
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You're a bit sloppy with your parentheses, but the result looks good if you make those corrections, replace your ε with ε0 and wind up with C = {ε0κ1κ2a2/D(κ12)} ln(κ1/κ2).
Note that I changed distance from d to D to avoid confusion with the calculus.
It's obvious you went the right way, good going.
 
carlyn medona said:
Okay I got

dc = εa^2k1k2dx ÷(d (xk1 -xk2+k2a) and after integration c
C = εa^2k1k2ln(k1÷k2)÷(d(k1-k2)
As rude man said, replace ε with ε0.

Just for fun, see what happens when you put k1=k2 in that equation.
 
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cnh1995 said:
Just for fun, see what happens when you put k1=k2 in that equation.
Good one, cnh! I wondered about that myself, thought at first I had made an error, finally determined the limit existed and agreed with the degenerate case k1=k2=k.
 
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