electromagnetism - Voltage on a Transmission Line with Ideal Conductors

Let's consider a transmission line with Perfect Electric Conductors.

We know that if an external AC source is applied, we get a voltage waveform between the conductors which is function of the position (and also of time, but focus on the first dependence).

But we know that in a perfect electric conductor the electric field is orthogonal to its surface, and this means that its surface is equipotential. This property is true in any situation (steady state or not), because the tangential electric field is always 0 in a perfect electric conductor.

The following picture shows clearly that E is orthogonal to the conductors' surface. enter link description here

But this seems in contrast with the fact that the voltage depends on the position.

Which is the solution?

Answer

The electric field being perpendicular to the conductor surfaces is a boundary condition imposed by their perfect conductance. The electric field lines terminate in the surface charge density according to

$$n \cdot \varepsilon \mathbf E = \sigma \space\space(1)$$ with n the normal vector to the surface, and σ the surface charge density. The question is now how a potential gradient can exist along the conductor surface while the electric field strength is normally incident.

A give-away is one of Maxwell's equations (Faraday's law):

$$\nabla × \mathbf E + \frac{\partial \mathbf B}{\partial t} = 0\space\space(2)$$ which essentially tells us that the electric field is non-conservative in the presence of a time-varying magnetic field. Let's study this a bit more in depth.

From vector calculus we know that the divergence of a curl and the curl of a gradient are both zero:

$$\nabla \cdot (\nabla × \mathbf A) = 0\space\space(3)$$ $$\nabla × (\nabla V) = 0\space\space(4)$$

As far as we know there is no magnetic charge, so according to Gauss' law for magnetic fields the magnetic induction has zero divergence:

$$\nabla \cdot \mathbf B =0$$ which allows us to use the identity of (3) to define a vector A such that $$\mathbf B = \nabla × \mathbf A\space\space(5)$$ We call A the magnetic vector potential. Substituting (5) in (2) gives $$\nabla ×( \mathbf E + \frac{\partial \mathbf A}{\partial t}) = 0$$ Now the identify of (4) allows us again to define the non-static electric scalar potential

$$\mathbf E + \frac{\partial \mathbf A}{\partial t} = -\nabla \varphi$$ which shows that a potential gradient can exist that does not have the same direction as the electric field vector, as long as there is a time varying magnetic vector potential, so there really is no contradiction that the potential gradient can be at angles with the electric field.