Capacitor with two dielectrics inserted diagonally

[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

 

[mfb]

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.

 

[Abhishek kumar]

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

 

[rude man]

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!

 

[cnh1995]

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.

 

[carlyn medona]

Okay I got

dc = εa^2k1k2dx ÷(d (xk1 -xk2+k2a) and after integration c
C = εa^2k1k2ln(k1÷k2)÷(d(k1-k2)

 

[rude man]

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(κ₁-κ₂)} 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.

 

[cnh1995]

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.

 

[rude man]

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.