Electromagnetic Fields and Waves: Notes, Solved Problems and Interactive Lab

Exercise 3.4.3

About the further exercises of Section 3.4

The ten exercises below are in the spirit of Problem 3.4.1, the home exercises and the worked examples. They use only $D = ε_{0} E + P = ε E$ , Gauss's law $\oint_{S} D \cdot d S = Q_{free}$ , the boundary conditions, the bound charges, $C = Q / V$ , $W = \frac{1}{2} \int D \cdot E d v$ and the image principle.

Exercise 3.4.3

Coaxial cable with graded insulation

cf. the coaxial capacitor

A coaxial cable has $a = 5$ mm and $b = 15$ mm. The insulation consists of two layers: $ε_{r 1} = 4$ for $a < ρ < c = 8$ mm and $ε_{r 2} = 2.5$ for $c < ρ < b$ . The highest permissible field in both materials is $10$ MV/m. Find (a) $C^{'}$ , (b) the field in both layers, and (c) the maximum voltage, compared with a single-layer insulation.

Coaxial cable cross-section with inner radius a, interface radius c and outer radius b, two insulation layers epsilon1 and epsilon2

Try it yourself first.

Strategy

Follow the four-step capacitance recipe with $\pm ρ_{L}$ per metre: Gauss's law gives the same $D_{ρ}$ in both layers, $E = D / ε$ differs, and the voltage is the sum of the two layer integrals. For the voltage rating, find where the field is largest in each layer (at its inner radius) and set the larger of the two equal to the limit.

(a)

Gauss's law for $D$ on a coaxial cylinder of radius $ρ$ and length $L$ gives $D_{ρ} 2 π ρ L = ρ_{L} L$ in both layers, so

E_{1} E_{2} = \frac{ρ _{L}}{2 π ε _{1} ρ} (a < ρ < c), = \frac{ρ _{L}}{2 π ε _{2} ρ} (c < ρ < b) .

The voltage is

V = \int_{a}^{c} E_{1} d ρ + \int_{c}^{b} E_{2} d ρ = \frac{ρ _{L}}{2 π} [\frac{ln ( c / a )}{ε _{1}} + \frac{ln ( b / c )}{ε _{2}}],

Answer

C^{'} = \frac{ρ _{L}}{V} = \frac{2 π ε _{0}}{\frac{ln ( c / a )}{ε _{r 1}} + \frac{ln ( b / c )}{ε _{r 2}}} = \frac{5.556 \times 1 0 ^{- 11}}{0.1175 + 0.2514} = 151 pF/m

The two layers act as two coaxial capacitors in series (Home exercise 3.4.1).

(b)

In each layer the field is largest at its inner radius:

E_{1} (a) = \frac{ρ _{L}}{2 π ε _{1} a}, E_{2} (c) = \frac{ρ _{L}}{2 π ε _{2} c}, \frac{E _{2} ( c )}{E _{1} ( a )} = \frac{ε _{1} a}{ε _{2} c} = \frac{4 \times 5}{2.5 \times 8} = 1.

The two maxima are equal: the layers are equally stressed.

(c)

With $E_{1} (a) = E_{m a x} = 10$ MV/m, $ρ_{L} = 2 π ε_{1} a E_{m a x} = 11.1$ μC/m and

V_{m a x} = E_{m a x} a [ln \frac{c}{a} + \frac{ε _{1}}{ε _{2}} ln \frac{b}{c}] = 1 0^{7} \times 5 \times 1 0^{- 3} [0.470 + 1.6 \times 0.629] = 73.8 kV .

With a single material throughout (either one), $V_{m a x} = E_{m a x} a ln (b / a) = 54.9$ kV.

Answer

C^{'} = 151 pF/m, V_{m a x} = 73.8 kV (against 54.9 kV for a single layer)

Exercise 3.4.3: the field in the cable at the same voltage V=73.8 kV. With graded insulation (orange) the field reaches the 10 MV/m limit twice, at \rho=a and at \rho=c, and stays below it elsewhere. With one material throughout (blue, dashed) the field V/(\rho\ln(b/a)) would reach 13.4 MV/m at the inner conductor, above the limit. — **Figure 3.4.32.** Exercise 3.4.3: the field in the cable at the same voltage $V = 73.8$ kV. With graded insulation (orange) the field reaches the $10$ MV/m limit twice, at $ρ = a$ and at $ρ = c$ , and stays below it elsewhere. With one material throughout (blue, dashed) the field $V / (ρ ln (b / a))$ would reach $13.4$ MV/m at the inner conductor, above the limit.

Check

$V_{m a x}$ for the graded cable must lie above the single-layer value because both layers are now used at the limit (Figure 3.4.32). Indeed, $73.8/54.9 = 1.34$ .

Remark

In a single material $E \propto 1/ ρ$ , and only the insulation near the inner conductor works at the permitted field. Putting the higher permittivity inside lowers $E$ where $ρ$ is small. With $ε_{1} a = ε_{2} c$ both layers reach $E_{m a x}$ together, and the same cable carries $34 %$ more voltage. This is the grading of high-voltage cables and bushings.

Exercise 3.4.3: Coaxial cable with graded insulation