Capacitance 2: Coaxial Transmission Line

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Coaxial transmission line

One of the most useful and common transmission lines is the coaxial line. Figure 1 shows the geometry, which consists of a cylindrical central conductor surrounded by a conductive cylindrical shield. Here, we shall describe how to find the capacitance per unit length of this line by using the the translational and rotational symmetries to simplify the problem in cylindrical coordinates.

Illustration of a circular coaxial transmission line. Potential is fixed on the center conductor and the conducting shield.
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Illustration of a circular coaxial transmission line. Potential is fixed on the center conductor and the conducting shield.

We know that the region of space between the conductors satisfies the Laplace equation, because there are no charges present. In cylindrical coordinates, this means

\nabla^2\Phi = \frac{1}{r}\frac{\partial}{\partial r}\left ( r\frac{\partial \Phi}{\partial r}\right ) + \frac{1}{r^2} \frac{\partial^2\Phi}{\partial \phi^2} + \frac{\partial^2\Phi}{\partial z^2} = 0

Rotational symmetry in the φ coordinate and translation symmetry in z mean that there is no variation in potential with respect to the azimuthal or longitudinal variables z and φ. we are then left with

\frac{1}{r}\frac{\partial}{\partial r}\left ( r\frac{\partial \Phi}{\partial r}\right ) = 0

This differential equation has the solution

\Phi = C_1 \ln r + C_2 \,,

where C_1, C_2\, are constants to be found using the boundary conditions present on the center conductor and the shield.

On the center conductor (r=a), Φ = 1. This means that

1 = C_1 \ln a + C_2\,.

On the outer shield, we have \Phi = 0\,, which gives

0 = C_1 \ln b + C_2\,.

We can now solve for C1 and C2:

C_1 = \frac{1}{\ln a - \ln b}\,
C_2 = \frac{-\ln b}{\ln a - \ln b}\,.

Rearranging and using the properties of logarithms, we find the final expression for the potential within the coaxial conductor as

\Phi(r) = \frac{\ln\frac{r}{b}}{\ln\frac{a}{b}}

The electric field is purely radial and equals

E_r(r) = -\frac{1}{\ln\frac{a}{b}}\frac{1}{r}

The charge per unit length present on the center conductor is proportional to the perpendicular electric field present at the surface of the conductor (found using Gauss's Law), i.e.:

Q/l = \frac{2\pi\epsilon}{\ln\frac{b}{a}},

where \epsilon\, is the dielectric permittivity of the insulating material filling the region between the conductors.

The capacitance per unit length is then found to be

C/l = \frac{Q/l}{1} = \frac{2\pi\epsilon}{\ln\frac{b}{a}},

since we chose the potential difference between conductors to be 1 at the start. (Although this in not important in capacitance calculations. The capacitance does not depend at all on the choice of voltage.)

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