I know that at a given frequency and therefore wavelength, a dipole antenna that is too (electrically) short looks capacitive, and one that is too long looks inductive. I can see it on the Smith Chart, and I can counteract it e.g. by "loading" a short antenna with an inductor. There are also some thought experiments out there, e.g. "0 length dipoles", that show why a short antenna must be capacitive to make sense.

But what I don't understand is what actually happens to the electromagnetic field to make the antenna look like an inductor or capacitor at the given wavelength.

I do (roughly, at least) know how current is distributed along a mismatched dipole: On a short dipole, the current is roughly triangularly distributed, because as you get closer to the extreme slope of the sine lobe the slope looks more and more linear. On a long dipole, you get larger and larger (and then more and more) opposite sign current lobes that cancel the electric field.

But how that makes the dipole effectively a capacitor or an inductor, I don't understand. I guess there must be some analogy on how the fields look and/or behave to how they do on either component. Is there some intuitive, maybe even visual explanation?

  • \$\begingroup\$ Excuse me, but you cannot see it on Smith Chart. That chart presents the behaviour of a TEM waveform transmission line which radiates nothing. Antennas generate totally different waveforms to the space. It's a common trick to make calculations for antennas as lossy transmission lines where the radiation is taken as a loss, but using Smith Chart essentially assumes the behaviour that you said you found. \$\endgroup\$
    – user136077
    Commented Jul 15, 2022 at 21:25
  • \$\begingroup\$ @user287001 I think that was my point. I can see the point on the Smith Chart move to be more inductive or more capacitive when adding or removing length (or am I wrong?), but it does not help understand nor explain. I was merely stating that I can't do that using the Smith Chart. \$\endgroup\$
    – anyfoo
    Commented Jul 15, 2022 at 21:38
  • \$\begingroup\$ If you want to see "something", go to this site amanogawa.com linear antennas menu ... \$\endgroup\$
    – Antonio51
    Commented Jul 16, 2022 at 16:36
  • \$\begingroup\$ If you think that a "short antenna" is an "open circuit", one can understand that it is a capacitor ... \$\endgroup\$
    – Antonio51
    Commented Jul 16, 2022 at 16:45
  • \$\begingroup\$ n4djantennaramblings.blogspot.com/2010/02/… \$\endgroup\$
    – RemyHx
    Commented Oct 26, 2022 at 5:00

1 Answer 1


This is a good question, but we can only see its "field related" answer in case of infinitesimal dipoles only.

In short, using the full field expressions (in the near region) of the infinitesimal dipole, we can derive the complex power propagating radially from the source using:

$$ S = \frac{1}{2} \int \int{(\vec{E} \times \vec{H}^*) . d\vec{A}} $$ This is from the complex Poynting theory is: $$ S = P_{rad} + 2j\omega(W_m-W_e) $$ In case of infinitesimal dipole, the imaginary part is negative, indicating more average electric energy than magnetic, exactly like a capacitive reactance.

Without knowing the field expressions in the near region, it is not possible to calculate the imaginary part of S. So, we can use that "field point of view" in case of infinitesimal dipoles only, where we can derive these field expressions in the near region. Otherwise, we look into the antenna input impedance obtained from measurements, simulations... or whatever.


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