Question

Could someone help to understand the derivation for capacitance and resistance of an coaxial cable?

Answer 1

The cylindrical Gaussian surface coaxial with the inner conducting cable, with length l and radius r. By Gauss law, we have

Electric field E = density (D) / 2πЄr where Є is dielectric constant

Voltage V = (density(D)/2πЄ) (ln r1/r2)
where r1 is inner conductor radius and r2 is outer conductor radius

Capacitance C =Charge Density (D) / Voltage

C = 2πЄ/(ln r1/r2)

Answer 2

\[E=E(r)=E_r=-dV(r,\theta)/dr \]
and
\[V=V(r)=Vr\]
then
\[E_r=-dV(r)/dr\]
\[V(r) = - \int\limits\limits_{r_1}^{r}E(u).du=-\int\limits_{r_1}^{r}C/u.du=C.\ln (r_1/r_2)\]
C is a constant see above for the value.

Answer 3

hmmmm....yeah i understand it now thanks but i saw another question asking to derive the formulas but they want it in the form \[C = L/2k \ln(b/a) \] how did they get the 2k??