Coaxial cable and Potential between

1. Oct 13, 2015

Bugge

Hello. I have DC going through a coaxial cable, and I have calculated the E fields of the two dielectrics in between to be $E_1$ and $E_2$ with help of their $D$-vectors. The dielectrics are cylindrically shaped like the conductors. As in, one is in contact with the inner conductor, and one is in contact with the outer.

Both fields vary by the distance $r$ through the $D$-fields,

$$a < r < c$$
$$c < r < b$$
Where $a$ is the inner conductor radius and $b$ is the outer conductor radius, and $c$ is inbetween the dielectrics.

Now I am not sure how to determine the potential. between them. Considering the different E-fields, how is,
$$V(r) = \int_b^a E d \mathcal{l}$$
expressed in my case? Would it be similar to
$$\int_a^b (E_2 - E_1) dr = \int_a^b E_2 dr - \int_a^b E_1 dr$$

I know the $E_n$-fields, and $J_n$- and $D_n$-vectors of the dielectrics. I also know the $\sigma_n$ of each of the dielectrics and the surface charge density of the inner ($\rho_si$) conductor and there is also a $\rho_sc$ between the dielectrics at distance $c$.

I'd appreciate any help, thank you very much!

2. Oct 14, 2015

davenn

Hi there
Im really tying to make sense of your cable makeup ?
I don't get your 2 dielectrics ... are both dielectrics between the inner and outer conductors ?
are these 2 dielectrics the same material ?
what separates the dielectrics ?
whatever it is must by definition be considered another (3rd) dielectric

draw a detailed diagram

Dave

3. Oct 15, 2015

Bugge

I could imagine it is not very descriptive, sorry. I found a link with a similar setup. The only difference is that there is a surface charge density $\rho_{s,c}$ in between the dielectrics at radius c (Nothing else is known of the boundary at radius c),
http://www-h.eng.cam.ac.uk/help/mjg17/teach/1AElectromag/coax.html

And I think the above link also answered my question (How is the potential found when there are two dielectrics in a coaxial cable).
The potential $V$ between the conductors are, as I can see, a sum of the potentials from the start to end of each dielectric boundary,

$$V = V_1 + V_2 = \int_a^c E_1 dl + \int_c^b E_2 dl$$

So you will have two integrals to sum up. I originally thought it would be the difference between the potentials, but summing them up do make sense to me physically.

Last edited: Oct 15, 2015