Electron Redistribution and Self-Repulsion in Antennas

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When a 10-volt peak-to-peak square wave is applied to an antenna, electrons flow in one direction during the positive voltage and in the opposite direction during the negative voltage. When the voltage drops to zero, the electrons reverse direction due to self-repulsion, as they return to their original state after being displaced. This behavior is influenced by the slope of the voltage change rather than the voltage sign itself. The discussion highlights the differences between antenna and capacitor behavior, emphasizing that the underlying physics of charge movement is consistent across both scenarios. The explanation of electron movement during the zero-voltage period is accepted as valid until proven otherwise.
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Let me rephrase my question here, since some of the responses in the orignal thread arent really what I am after, maybe this is more of a physics question than an electrical engineering one...

If we apply a 10 volt peak to peak square wave to an antenna (assume the signal goes from +5 volts to -5 volts) we have the following situation...

During the time we apply the +5 volts the electrons in the antenna "flow" in one direction and during the time we apply the -5 volts the electrons in the antenna "flow" in the opposite direction.


Now we apply a 10 volt peak to peak square wave that will range from 0 volts to + 10 volts. What happens to the electrons in the antenna in this case?

During the time we apply the +10 volts I can see the electrons "flowing" in the same direction as they did when we applied +5 volts. But when we change the applied voltage from +10 volts to 0 volts I am not sure why they would flow in the opposite direction since we arent really applying a negative voltage, we are applying no voltage at all.

Does this occur because during the time we apply +10 volts the electrons become "displaced" in one direction and then when we drop the voltage dwon to zero the electrons return to the original state due to the electromagnetic repulsion between electrons (before we applied any voltage)?

Is this a good explanation of why the electrons would move in the opposite direction during the period of no applied voltage?




Russ
 
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steadele said:
If we apply a 10 volt peak to peak square wave to an antenna (assume the signal goes from +5 volts to -5 volts) we have the following situation...

During the time we apply the +5 volts the electrons in the antenna "flow" in one direction and during the time we apply the -5 volts the electrons in the antenna "flow" in the opposite direction.

taking an antenna makes things more difficult, is this necessary ?

Moreover, on any impedance different from a purely resistive one, you have a phase difference between the voltage and the current. The voltage might be positive, and nevertheless, the current might flow out of the load and into your generator.

The impedance of an antenna depends on the frequency used, its structure...

So let us rather take a capacitor. You're already in for a surprise btw.

When the voltage rises, electrons will flow out of the capacitor into your generator. When the voltage is constant, nothing will move. When the voltage will lower, electrons will flow from the generator into the capacitor. It doesn't matter what sign the voltage has, it is the slope that counts. That's why the equation for a capacitor is:
i = C dv/dt.
 
vanesch said:
taking an antenna makes things more difficult, is this necessary ?

Yes it is necessary because I am looking at what happens in the two cases I describe above.

Im looking for the cause of the electrons reversing direction when we apply zero volts.

vanesch said:
Moreover, on any impedance different from a purely resistive one, you have a phase difference between the voltage and the current. The voltage might be positive, and nevertheless, the current might flow out of the load and into your generator.

The impedance of an antenna depends on the frequency used, its structure...

Yes, I know.

vanesch said:
So let us rather take a capacitor. You're already in for a surprise btw.

Actually nothing you wrote surprises me. I already knew about electron flow in a capacitive circuit.

vanesch said:
When the voltage rises, electrons will flow out of the capacitor into your generator. When the voltage is constant, nothing will move. When the voltage will lower, electrons will flow from the generator into the capacitor. It doesn't matter what sign the voltage has, it is the slope that counts. That's why the equation for a capacitor is:
i = C dv/dt.

Yes I know this. If you charge up a capacitor to a certian source voltage in a circuit and then reduce the source voltage the stored charge in the capacitor will cause charge to flow in the opposite direction (opposite to the direction that it was flowing during the charging cycle).

But a capacitor is not the same thing as an antenna (although there are similarities I will admit).


As for my antenna example given in the first post...

Can we say that the reason the electrons will move in the opposite direction during the 0 volt time period is because they are returning to the original state in the wire before we applied the 10 volts?

Can we also say that, since 0 volts implies no voltage, that this reversal of direction is caused by the repulsion of the electrons (which have been slightly displaced to one end of the antenna due to the +10 volts we applied a moment ago) with one another?

This sounds like a reasonable physical explanation to me. Thoughts?





Russ
 
steadele said:
Yes it is necessary because I am looking at what happens in the two cases I describe above.

Im looking for the cause of the electrons reversing direction when we apply zero volts.

Self-repulsion. It's what causes charges on a surface to redistribute themselves when there are no external field.

Zz.
 
ZapperZ said:
Self-repulsion. It's what causes charges on a surface to redistribute themselves when there are no external field.

Zz.


Thank you. I agree that this must be the cause. Its nice to have someone confirm this.

I will accept this explanation as correct until someone can prove it wrong.




Russ
 
Hello! Let's say I have a cavity resonant at 10 GHz with a Q factor of 1000. Given the Lorentzian shape of the cavity, I can also drive the cavity at, say 100 MHz. Of course the response will be very very weak, but non-zero given that the Loretzian shape never really reaches zero. I am trying to understand how are the magnetic and electric field distributions of the field at 100 MHz relative to the ones at 10 GHz? In particular, if inside the cavity I have some structure, such as 2 plates...
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