---
title: "Dielectric stress maximum at conductor surface"
subject: "Power System Analysis I"
chapter: 2
key: "semester_4/power_system_analysis_i:dielectric-stress-max-surface"
source: https://www.bctengineer.com/q/semester_4/power_system_analysis_i/dielectric-stress-max-surface
site: BCT Engineering
---

# Dielectric stress maximum at conductor surface

Power System Analysis I · Chapter 2: Overhead and Underground Transmission

- Asked in: 2077 Chaitra, 5 marks; 2075 Baisakh, 6 marks (asked 2 times)
- Video: [Dielectric Stress of a Single Core Cable || Power System Analysis](https://www.youtube.com/watch?v=St6mEWzDpag) — Rojib EEE Academy
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/power_system_analysis_i/dielectric-stress-max-surface
- Whole chapter: https://www.bctengineer.com/md/semester_4/power_system_analysis_i/chapter-2

## Question

Prove that dielectric stress is maximum at the conductor surface and its value goes on decreasing as we move away from the conductor.

## Short answer

In a single-core cable, dielectric stress gx = V/(x ln(R/r)), inversely proportional to radius. Maximum stress occurs at the conductor surface (x=r) and minimum at the sheath (x=R). Example: gmax ≈ 36 kV/cm for V=33 kV, r=1 cm, R=2.5 cm.

## Answer

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.
- $\dfrac{g_{max}}{g_{min}} = \dfrac{R}{r}$.

```text
 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 $R/r = e = 2.718$.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. In a single-core cable, how does the dielectric stress g_x vary with the radial distance x from the axis?
   A. Directly with x
   B. Inversely with x, g_x = V/(x ln(R/r))
   C. Inversely with x^2
   D. It is constant across the insulation
   Answer: B. Gauss's law gives g_x = Q/(2 pi eps x), which is proportional to 1/x. (Key point: Stress varies as 1/x)

2. Where is the dielectric stress maximum and what is its value?
   A. At the sheath, g_max = V/(R ln(R/r))
   B. Midway in the insulation, at x = (r+R)/2
   C. At the conductor surface, g_max = V ln(R/r)/r
   D. At the conductor surface, g_max = V/(r ln(R/r))
   Answer: D. Since g is proportional to 1/x, it is largest at the smallest x, i.e. x = r. (Key point: Location of maximum stress)

3. How is the charge term Q/(2 pi eps) eliminated to express g_x in terms of the applied voltage?
   A. Integrate g_x from 0 to R
   B. Set Q equal to V
   C. Integrate g_x from r to R to get V = Q/(2 pi eps) ln(R/r)
   D. Differentiate g_x with respect to x
   Answer: C. The potential difference V is the integral of g_x dx between conductor and sheath, giving Q/(2 pi eps) = V/ln(R/r). (Key point: Using V to eliminate Q)

4. What is the ratio g_max / g_min in the cable insulation?
   A. r/R
   B. ln(R/r)
   C. (R/r)^2
   D. R/r
   Answer: D. g_max/g_min = (1/r)/(1/R) = R/r. (Key point: Max to min ratio)

## Related in Power System Analysis I

- [Optimum conductor radius for minimum cable stress](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Acable-optimum-conductor-radius) — Power System Analysis I, chapter 2

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