[Electricity and magnetism] For a half-wavelength dipole antenna, why is the voltage distribution a quarter wavelength out of phase with the current?

Submitted by VainVeinyVane t3_10m5c38 in askscience

Solved! - See my comment in the replies.

I'm talking about this photo:

https://en.wikipedia.org/wiki/File:Dipole_antenna_standing_waves_animation_6_-_5fps.gif

As far as I know, voltage and current are in phase at the feed point. So when they form a standing wave, shouldn't they form standing waves that are in phase with each other? Why is the standing wave for the voltage a quarter wavelength out of phase with the current?

74

Comments

You must log in or register to comment.

neffs t1_j61r4m8 wrote

Read the last part of the description of the gif;

Dipoles have relatively high Q factor so the amount of energy stored in the standing waves is large compared to the energy added each cycle by the feedline, the feed voltage just represents a small perturbation to the standing waves. This is why the voltage standing wave is much larger than the voltage step at the feedline. Since the standing waves are storing energy, not transporting power, the current in them is not in phase with the voltage but 90° out of phase.

"Q Factor"

13

VainVeinyVane OP t1_j641t7v wrote

OK. Got the answer. For anybody who stumbles upon this question and has the same question: Current is proportional to direction and concentration of charge, whereas Voltage is only proportional to concentration of charge.

So when the electrons "bounce" off the end of the end of the antenna (the wave is reflected), the NEGATIVE of the wave is reflected for current (since the direction reverses!), while the POSITIVE of the wave is reflected for the voltage.

For example:

if the wave is cos(wt-kx), the reflected traveling wave is -cos(wt+kx) for current, while the reflected wave for voltage is +cos(wt+kx).) The resulting standing waves formed are one with peaks at the ends for voltage, while the current's resulting standing wave has its peak at the center and nodes at the ends.

​

The graph for current, cos(wt-kx) - cos(wt+kx) will be 2sin(wt)sin(t). This is a standing wave with nodes at the ends of the antenna.

However, the graph for voltage, cos(wt-kx)+cos(wt+kx) will be 2cos(wt)cos(kx). This is a standing wave with the node at the center of the antenna and peaks at the ends (90 degrees shifted from the current standing wave).

​

Hope this helps!

2

zutnoq t1_j656wje wrote

Replying to VainVeinyVane (#1,529,820)

If you want a more hand wavy explanation I think the antenna is essentially acting as a giant capacitor (or rather one side of a capacitor) which would certainly explain the 90° phase difference between voltage and current as that is exactly what happens with a regular capacitor driven with AC (the exact opposite phase shift happens for an ideal inductor driven with AC).

4

zutnoq t1_j67q8kj wrote

Replying to VainVeinyVane (#1,539,580)

A capacitor is essentially two plates of metal separated by a layer of insulator. The antenna is acting like one of those plates of metal, the air is the insulator and (the) ground is acting as the other plate.

Edit: in this specific case the other arm of the antenna is the other plate, if that wasn't obvious.

2