Optimum conductor radius for minimum cable stress

In a single-core cable, the electric stress is greatest at the conductor surface. For a fixed sheath radius, this maximum stress is lowest when $R/r = e$.

Stress in a single-core cable

Let the conductor radius be $r$, the inside radius of the sheath $R$, and the voltage between core and sheath $V$. The charge per metre is $q$. By Gauss's law, the field at radius $x$ is:

$$g_x = \frac{q}{2\pi\varepsilon_0\varepsilon_r x}$$

Integrating from $r$ to $R$:

$$V = \frac{q}{2\pi\varepsilon_0\varepsilon_r}\ln\frac{R}{r} \;\Rightarrow\; g_x = \frac{V}{x\ln(R/r)}$$

The stress is largest at $x = r$:

$$g_{max} = \frac{V}{r\ln(R/r)}$$

Condition for minimum $g_{max}$

With $V$ and $R$ fixed, $g_{max}$ is least when the denominator $f(r) = r\ln(R/r)$ is greatest:

$$\begin{aligned} \frac{df}{dr} &= \ln\frac{R}{r} + r\cdot\left(-\frac{1}{r}\right) = \ln\frac{R}{r} - 1 = 0\\ \ln\frac{R}{r} &= 1 \;\Rightarrow\; \frac{R}{r} = e = 2.718 \end{aligned}$$

The second derivative, $d^2f/dr^2 = -1/r$, is negative, so this point is a maximum of $f$ and therefore a minimum of $g_{max}$.

Result:

$$r = \frac{R}{e}, \qquad g_{max}\big|_{min} = \frac{V}{r} = \frac{eV}{R}$$

The insulation thickness is then $R - r = r(e-1) = 1.718r$.

Practical note

The conductor radius from $r = R/e$ is often larger than the radius needed to carry the current. In that case a hollow conductor or a larger-diameter (for example aluminium) conductor is used to obtain the optimum radius.

Example

Take $V = 66$ kV (core to sheath) and $R = 2$ cm. Then $r = 2/2.718 = 0.736$ cm and $g_{max} = 66/0.736 = 89.7$ kV/cm.