How do I calculate the induced voltage on a finite antenna

Question

I'm trying to calculate (numerically) the induced voltage on a half-wave dipole caused by a moving charge as a function of time. Up to this point the way I have been doing this is using the formula

$V = \vec{E} \cdot \vec{h} \,,$

where $\vec{E}$ is the electric field from the moving charge, measured at the centre point of the antenna and $\vec{h}$ is the effective length of the antenna. The literature informs me that

$\vec{h}(\theta) = h(\theta) \, \hat{\theta} = \frac{\lambda}{\pi} \frac{\cos(\frac{\pi}{2}\cos\theta)}{\sin\theta} \hat\theta $

where $\theta$ is the angle between the axis of the dipole and the radiating charge.

My question is, how much of an approximation is this? My antenna has a non-zero size and thus the electric field (which is not sinusoidal in time) is going to vary as a function of time and space along the antenna due the propagation time of the EM waves. Is there an integral over the length of the antenna (or alternatively in time) that needs to be performed to get the true voltage? If so, what form does this take?

I suspect that any correction to the current case is going to be fairly minor from my own tests of plotting the electric field at various points along the antenna but it would be good to confirm this since I would like to simulate an arbitrary antenna at some point.

Answer 1

The accelerated charge's electric field $\mathbf E_{charge}$ causes an external electromotive force in the antenna, which can be expressed as integral over the body of the antenna in the chosen positive direction of the antenna dipole:

$$ \mathscr{E} = \int_{antenna~rods} \mathbf E_{charge} \cdot d\mathbf s . $$

This is not (conceptually) the voltage = difference of potential between two antenna terminals yet, because the effect of the antenna's own field isn't taken into account. That depends on what the antenna terminals are connected to.

If they are not connected to anything (zero load), then induced current in the antenna will be very small and we can approximate and assume it is zero. Since a good antenna is made of a good conductor, zero current implies zero net electric field inside. So in this approximation, the external electric field is completely cancelled out by the antenna electric field due to charges in the antenna itself. Since there is almost no current, this antenna electric field has no induced field component (since there is almost no changing current), and is thus entirely due to electrostatic field of surface charges on the antenna. Only in this case, we can conclude that difference of potential is equal in magnitude to $\mathscr{E}$, and it is of course opposite in sign (so that the two field effects cancel each other out in the antenna conductor).

If the antenna terminals are connected to a conductive load (a low resistor, a large capacitor, or some appropriate circuit allowing current to flow), then current in the antenna may no longer be negligible, and induced electric field of the antenna may become important. This induced electric field helps in counteracting the external electric field in the antenna, so the induced difference of electric potential = voltage between the two antenna terminals may be lower in magnitude than $\mathscr{E}$. Finding the value can be a complicated problem depending on the accuracy needed. In the simplest model, one may treat the system as some RLC circuit powered by an emf source of some internal resistance. In a more detailed model, one would assume some definite geometry of the antenna (and possibly also the relevant parts of the connected circuit) and then for this fixed geometry, search for parameters of equivalent emf source (emf value and internal resistance) via solving some equivalent of Maxwell's equations for such a system, with help of the computer.