Capacitor with two dielectrics inserted diagonally

In summary, to find the capacitance of a capacitor with square plates of length a separated at a distance d and dielectrics inserted as shown in the figure, you can use the formula C = {ε0κ1κ2a2/D(κ1-κ2)} ln(κ1/κ2), where κ1 and κ2 represent the dielectric constants of the two materials. To simplify the integration process, it is easier to form columns of equal dielectric distribution and then evaluate one integral. The degenerate case of k1=k2 will result in a limit that exists and agrees with the formula.
  • #1
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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  • #2
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.
 
  • #3
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
 
  • #4
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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  • #5
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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  • #6
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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  • #7
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.
 
  • #8
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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  • #9
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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1. What is a capacitor with two dielectrics inserted diagonally?

A capacitor with two dielectrics inserted diagonally is a type of capacitor where two different dielectric materials are used to separate the two conducting plates, with one dielectric material filling one half of the space between the plates and the other dielectric material filling the other half.

2. How does a capacitor with two dielectrics inserted diagonally differ from a regular capacitor?

A regular capacitor has only one dielectric material separating the two plates, while a capacitor with two dielectrics inserted diagonally has two different dielectric materials inserted in a diagonal pattern. This allows for a higher capacitance value and better insulation properties.

3. What is the purpose of using two dielectrics in a capacitor?

The use of two dielectrics in a capacitor allows for a larger capacitance value by increasing the effective area between the plates. It also helps to reduce the overall size and weight of the capacitor while maintaining the same capacitance value.

4. How does the dielectric constant of each material affect the overall capacitance of the capacitor?

The dielectric constant of a material is a measure of its ability to store electrical energy. A higher dielectric constant results in a higher capacitance value, so using two dielectrics with different dielectric constants can increase the overall capacitance of the capacitor.

5. What are some common applications of a capacitor with two dielectrics inserted diagonally?

A capacitor with two dielectrics inserted diagonally is commonly used in electronic circuits, power supplies, and high-frequency applications. It is also used in energy storage devices, such as in hybrid and electric vehicles, due to its high capacitance and compact size.

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