Fernando Garcia

PHYS405 - Electricity & Magnetism I

A certain coaxial cable consists of a copper wire, radius $a$ , surrounded by a concentric copper tube of inner radius $c$ (Fig. 4.29).

The space between is partially filled (from $b$ out to $c$ ) with material of dielectric constant $ϵ_{r}$ , as shown.

Find the capacitance per unit length of this cable.

Here's a more explicit version of Figure 4.29

Where it is worth remarking that copper is a conductor.

Using Gauss's law for $D$ , we see that (cylindrical Gaussian surface)

D \cdot (area) = D (2 π s ℓ) = Q

Where $Q$ is some charge on the cable.

So $D$ is

D = \frac{Q}{2 π s ℓ}

Now that we know $D$ , we can use it to find $E$ :

E = \frac{1}{ϵ} D

Where

ϵ = ϵ_{0} (1 + χ_{e})

Up until $s = b$ (from $s = a$ ), $χ_{e} = 0$ so:

\begin{aligned} E_{a \leq s \leq b} & = \frac{1}{ϵ_{0}} D \\ = \frac{Q}{2 ϵ_{0} π s ℓ} \hat{s} \end{aligned}

But once $s > b$ , we enter a dielectric so $χ_{e} \neq 0$ and here

\begin{aligned} E_{b \leq s \leq c} & = \frac{1}{ϵ} D \\ = \frac{Q}{2 ϵ π s ℓ} \hat{s} \end{aligned}

For some $ϵ$ given by $χ_{e}$ depending on the material.

We want the capacitance (per unit length), so let's first find the potential inside:

\begin{aligned} V & = - \int_{c}^{a} E \cdot d l \\ = - \int_{c}^{b} \frac{Q}{2 π ϵ ℓ} \frac{1}{s} d s - \int_{b}^{a} \frac{Q}{2 π ϵ_{0} ℓ} \frac{1}{s} d s \\ = \frac{Q}{2 π ϵ_{0} ℓ} (\ln (b / a) + \frac{ϵ_{0}}{ϵ} \ln (c / b)) \\ = \frac{Q}{2 π ϵ_{0} ℓ} (\ln (b / a) + \frac{1}{ϵ_{r}} \ln (c / b)) \end{aligned}

Capacitance per unit length is given by $C / ℓ$ , and $C$ is given by $C = Q / V$ , so:

\begin{aligned} capacitance per unit length \\ = \frac{C}{ℓ} \\ = \frac{1}{ℓ} \frac{Q}{V} \\ = \frac{Q}{ℓ} \frac{ℓ}{Q} 2 π ϵ_{0} \frac{1}{\ln (b / a) + \frac{1}{ϵ_{r}} \ln (c / b)} \\ = \frac{2 π ϵ_{0}}{\ln (b / a) + \frac{1}{ϵ_{r}} \ln (c / b)} \end{aligned}