Answer to Question #181041 in Electricity and Magnetism for Kristi Meacham

Question

What is the capacitance and self-inductance per length of a cable that consists of a solid inner conductor (radius R₁) and a thin coaxial conducting cylindrical shell (inner radius R₂). The region between the conductors is filled with a uniformly distributed dielectric (dielectric constant Kappa). Mu=Mu₀

Answer 1

Capacitance 

Assume an imaginary Gaussian cylinder of radius r where $R_2 \eqslantless r\eqslantless R_1$

By Gauss law $\int \vec{E}.\vec{d}a= \frac{Q}{\epsilon} \implies |\vec{E}|. 2 \pi rh =\frac{Q}{\epsilon} \implies |\vec{E}|= \frac{Q}{\epsilon 2 \pi rh}$

Voltage difference integral of electric field , $V=\int _{R_2} ^{R_1} \vec{E}dr= \frac{Q}{\epsilon 2 \pi rh} \int _{R_2} ^{R_1} \frac{1}{r}dr = \frac{Q}{\epsilon 2 \pi h} \ln(\frac{R_1}{R_2})$

We know Q=CV

$Q=\frac{CQ}{\epsilon 2 \pi h} \ln(\frac{R_1}{R_2}) \implies \frac{C}{h}=\frac{\epsilon 2 \pi}{h} \ln(\frac{R_1}{R_2})$ h represents the length

Inductance 

Using Ampere's law and imaginary circular path of radius r

$\oint \vec{B}.\vec{d}l= \mu I \implies \vec{B}. 2 \pi r = \mu I \implies \vec{B}= \frac{ \mu I }{ 2 \pi r}$

If you consider a flux on the surface

$\phi=\int _{R_2} ^{R_1} \frac{\mu I}{2 \pi}hdr= \frac{\mu I h}{2 \pi} \int _{R_2} ^{R_1} \frac{1}{r}dr = \frac{\mu I h}{2 \pi} \ln(\frac{R_1}{R_2})$

We know $V_{drop}= \frac{d \phi}{dt}=L\frac{dI}{dt} =\frac{\mu h}{2 \pi} \ln(\frac{R_1}{R_2})\frac{d I}{dt}$

$\frac{L}{h}=\frac{\mu}{2 \pi} \ln(\frac{R_1}{R_2})\frac{d I}{dt}$ h represents the length