Voltage on a Transmission Line with Ideal Conductors

Question

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 1

Remember that when we defined the electrostatic potential difference (aka "voltage"),

$$V=-\int \vec{E}\cdot d\vec\ell,$$

we called it the electrostatic potential difference because it is only strictly valid in electrostatics. When we use this concept in AC circuits, we're using it as an approximation only (usually described as the lumped circuit approximation). In particular, in the presence of time-varying magnetic fields, we can't count on this \$V\$ to be independent of the path over which we take the integral.

In transmission lines, we are definitely dealing with time-varying magnetic fields, so we can't expect the electrostatic potential difference to be well defined.

We define an approximate potential at a point along the transmission line as the negative integral of the electric field from one conductor to the other at that point.

But we can't expect to get the same result (because this isn't an electrostatics situation) if we take an integral from some point on the first conductor, lengthwise along the conductor (with contribution 0, because the material is p.e.c.), across the gap at another location, and then back along the second conductor (again contributing 0) to the point opposite where we started.

Meaning, if we take the integral to calculate the "voltage" across the transmission line at position \$z=0\$, we can't expect this integral to be the same at \$z=z_1\$ just because there's no electric field within the conductors.

Answer 2

You’re driving with AC. You can’t ignore time-varying effects.

The line has a distributed capacitance: there’s different charge densities at points with different E fields and potentials.

Since the E and potential at each point varies with time, so does the charge: there are time-varying currents to change those charges.

Those in turn cause time varying B fields around the conductors.

And the additional E fields induced by those time-varying B fields re what you’re looking for.

Answer 3

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.

It's probably better to flip it around and say 'when the surface is equipotential, the electric field will be orthogonal to (normal to) its surface'.

In a transmission line, which you can model as having distributed inductance, no matter how perfect the conductor, the voltage along the inductor clearly varies with position if the transmission line is carrying anything other than DC. The surface is not equipotential, the electric field is not normal to the surface.

When dealing with transmission lines, AC makes my head hurt, I prefer to launch a step along it, and see the change of voltage along the line together with the change of current charging the line.

Answer 4

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.