effective permitivity of two cylinders with common center

[yefj]

Hello,i have two dielectric cylinders going one inside the other.
how do i calculate the effective permitivity of the whole scturcture?
Thanks.

 

[albbg]

If you mean a cylindrical capacitor with two different coaxial dielectics, the solution is quite simple. We know the capacitance of a cylindrical cap is given by: 2pi*eps/ln(R2/R1) where R1 and R2 are the inner and outer radius. If you have two coaxial dielectric, so the radii are R1, R2 and R3, you can consider the structure as two capacitor in parallel having the to capacitances per unit length:

C1 = 2*pi*eps1/ln(R2/R1)
C2 = 2*pi*eps2/ln(R3/R2)

1/Ceq = 1/C1 + 1/C2

cosidering now the two dielectric as one having equivalente permittivity eps_e and inner radius R1, outer radius R3 we will have:

Ceq =2*pi*eps_eq/ln(R3/R1)

that means

ln(R3/R1)/2*pi*eps_eq = ln(R2/R1)/2*pi*eps1 + ln(R3/R2)/2*pi*eps2

now, by means of simple math:

eps_e = eps1*eps2*ln(R3/R1) /[eps2*ln(R2/R1) + eps1*ln(R3/R2)]

 

[Easy peasy]

Permittivity refers to the medium - air - between the cylinders - measuring the capacitance allows calculation of permittivity for the (usually ) insulating medium.

 

[yefj]

two dielectric cylindres one inside the other,not capacitative cylinders.

 

[albbg]

Could you explain better ? Is this strutture placed inside an electric field ? Which direction with respect to the vector field ?

 

[yefj]

Hello albbg,the structure is shown bellow.
Thanks.

 

[albbg]

It's not isotropic. The permittivity depends on the orientantion and the shape of the electric field. How about the purpose of this structure ? Is used for what ?

 

[D.A.(Tony)Stewart]

Permittivity is measured in Farads/meter. (when in doubt, assume common sense)

For the axial direction, the effective dielectric constant, Dk (or relative permittivity) is the product*sum of each radial area*Dk divided by the total area.

Now if you wanted to compute that in the coaxial direction with two dielectrics in series, that which requires an infinitely small center conductor which is very lossy and inductive (bad idea) otherwise you need to add center hole dimension.

 

[albbg]

Permittivity is measured in Farads/meter. (when in doubt, assume common sense)

For the axial direction, the effective dielectric constant, Dk (or relative permittivity) is the product*sum of each radial area*Dk divided by the total area.

Now if you wanted to compute that in the coaxial direction with two dielectrics in series, that which requires an infinitely small center conductor which is very lossy and inductive (bad idea) otherwise you need to add center hole dimension.

In case of a capacitor with infinitely small center conductor, the permittivity will be eps1 regardless the dimensions (just take the limit of eps_e for R1-->0). However you can put the structure in an electric field and it will exhibit a certain permittivity, f.i. you can put it between two metallic plates. But the result will depend form the orientation of the structure switch respect to the lines of the electric field.

 

[D.A.(Tony)Stewart]

I agree you could do that, but I expect that is not what he meant from a simple question.