Dielectric stress maximum at conductor surface

In a single-core cable, the electric field (dielectric stress) in the insulation is inversely proportional to the radius, so it is highest at the conductor surface and falls towards the sheath.

Derivation

Let conductor radius $= r$, inner sheath radius $= R$, charge $= Q$ C/m length, permittivity $\varepsilon = \varepsilon_0\varepsilon_r$. Take a coaxial cylinder of radius $x$ ($r < x < R$) and length 1 m.

By Gauss's law, flux density and field at $x$:

$$D_x = \frac{Q}{2\pi x},\qquad g_x = \frac{D_x}{\varepsilon} = \frac{Q}{2\pi\varepsilon x}$$

Potential difference between conductor and sheath:

$$V = \int_r^R g_x\,dx = \frac{Q}{2\pi\varepsilon}\ln\frac{R}{r} \quad\Rightarrow\quad \frac{Q}{2\pi\varepsilon} = \frac{V}{\ln(R/r)}$$

Substituting:

$$g_x = \frac{V}{x\ln(R/r)}$$

Conclusion

Since $V$ and $\ln(R/r)$ are constants, $g_x \propto \dfrac{1}{x}$:

• At the conductor surface ($x = r$): $g_{max} = \dfrac{V}{r\ln(R/r)}$, the maximum.
• At the sheath ($x = R$): $g_{min} = \dfrac{V}{R\ln(R/r)}$, the minimum.
• $\dfrac{g_{max}}{g_{min}} = \dfrac{R}{r}$.

 stress g
  ^
  |*
  | *
  |   *
  |      *  *   *
  +--+-------------+---> x
     r             R

Example

$V = 33$ kV, $r = 1$ cm, $R = 2.5$ cm: $\ln 2.5 = 0.916$, $g_{max} = 33/(1\times0.916) = 36.0$ kV/cm, $g_{min} = 14.4$ kV/cm.

Because the inner insulation is most stressed, cables use grading (capacitance or intersheath grading) to make the stress more uniform; also the most economical size is when $R/r = e = 2.718$.