Dielectric Capacitance of Parallel Plate Cap

[jesuslovesu]

Homework Statement

A parallel plate cap of cross sectional area A and thickness d is filled with a dielectric material whose relative permittivity varies from $$\epsilon_r = 1 to \epsilon_r = 10$$
Find the capacitance.

Homework Equations

The Attempt at a Solution

C = Q/V

E = $$\sigma / e0 \epsilon_r$$

I am thinking that since V = $$- \int{}{} E dot dl$$ I need to integrate, I can't quite figure out the relationship between this though.

I kind of would like to just evaluate the from 10 to 1 and get C = 9 A e0 / d but that isn't quite it.
C = 9A e0 / d is close to the answer, but somehow a ln sneaks in there too?

 

[olgranpappy]

how does the dielectric vary from 1 to 10? linearly?

 

[olgranpappy]

Homework Statement

A parallel plate cap of cross sectional area A and thickness d is filled with a dielectric material whose relative permittivity varies from $$\epsilon_r = 1 to \epsilon_r = 10$$
Find the capacitance.

Homework Equations

The Attempt at a Solution

C = Q/V

E = $$\sigma / e0 \epsilon_r$$

I am thinking that since V = $$- \int{}{} E dot dl$$ I need to integrate

no
$$V=\int \vec E \cdot d\vec \ell=\int_{x=0}^{x=\ell}\frac{D}{\epsilon(x)}dx=D\int_{x=0}^{x=\ell}\frac{1}{\epsilon(x)}dx$$
and if eps varies linearly one does end up with a log term in the capacitance. (here D=\epsilon E is the electric displacemnt field)

I kind of would like to just evaluate the from 10 to 1 and get C = 9 A e0 / d but that isn't quite it.
C = 9A e0 / d is close to the answer, but somehow a ln sneaks in there too?[/QUOTE]

 

[jesuslovesu]

Yes, sorry it's linearly.


Cool thanks, I'll give it a shot now.