If a coaxial cable has a coaxial return conductor with infinite outer radius, will the return conductor experience a voltage build-up due to current flowing through it, or will it stay on ground potential? Here I'm taking infinite to be the zero potential.

I would appreciate if both the d.c. and a.c. case are discussed, if the answer is dependent on type of excitation.

Edit: I think the specific problem where my intuition lets me down is whether there is a scalar potential build-up axially along the return conductor, such that the electric field in the return conductor is given by

$$ \mathbf{E} = -\nabla \varphi - \frac{\partial \mathbf{A}}{\partial t} $$

or, if the electric field is only due to the changing magnetic field, i.e. $\mathbf{E} = - \frac{\partial \mathbf{A}}{\partial t}$.

My reason for thinking it's the latter is that since the return conductor certainly is a good conductor, then $\varphi$ must be constant and equal to zero, or else currents would flow radially to cancel the charge/potential build-up. Is this correct?

  • $\begingroup$ Draw a diagram. The term "return conductor" is obscure. $\endgroup$
    – DanielSank
    Jun 19, 2022 at 7:17

1 Answer 1


AC current, due to the skin effect, tends to flow on the inner surface of the outer conductor of a coax cable. The higher the frequency, the thinner the skin depth. For instance, at $1$MHz, most of the current will flow inside a layer of a couple of hundreds of microns.

DC current will spread out much more, but most of the current will flow within the radius comparable with several lengths of the cable, since the resistance of a path beyond that would substantially exceed the resistance of the direct path.

So, in both cases, the return current will cause some voltage drop, but, in the DC case, it would be much smaller.

  • $\begingroup$ Thank you! Could you please provide more details to your answer by taking into account the edit of my question? That is, the differentiation between electric field and scalar potential. $\endgroup$
    – electro_h
    Nov 15, 2018 at 14:00
  • $\begingroup$ @electro_h Obviously, for DC, there is no changing magnetic field, and therefore, no magnetically induced electric field. So, whatever negligibly small voltage drop is measured, it would be due to resistance. For AC, there is a changing magnetic field, but a coax cable is a transmission line and the actions of distributed inductance and distributed capacitance neutralize each other, resulting in the overall resistive characteristic impedance. So, for AC, electric field or voltage drop along the cable will be also due to resistance and will grow with frequency due to the skin effect. $\endgroup$
    – V.F.
    Nov 15, 2018 at 15:36
  • $\begingroup$ Your argument as to why current does not flow everywhere in the return conductor is due to the cable being of finite length. Neglecting end-effects, I think the current would use to whole return conductor. This means that $\mathbf{J} \to 0$ when $\omega \to 0$, which means that $\mathbf{E} = \rho \mathbf{J} \to 0$, which means that $\nabla \varphi = 0$. Also, your statement about coaxial cables having a resistive characteristic impedance is not correct in the general case. Thank you for a few clarifying ideas, but I can't accept your answer. $\endgroup$
    – electro_h
    Nov 21, 2018 at 7:24
  • $\begingroup$ @electro_h Fair enough. I would not argue about the DC case: the voltage drop would be negligible whether the radius is infinite or just several lengths of the cable. As for the AC, if you imply that the voltage drop along the length of the outer conductor has a significant non-resistive (say, inductive) component, I would like to hear your argument and see how this voltage drop is affected by frequency. $\endgroup$
    – V.F.
    Nov 21, 2018 at 21:23

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