Does half wave in dipole means that duration of 1/4 of a one signal cycle should equals capacitor charging duration, if we consider that dipole is a capacitor?

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  • $\begingroup$ First, why do you want to consider a dipole as a capacitor? $\endgroup$ – Jon Custer Aug 22 '18 at 14:27
  • $\begingroup$ @JonCuster, because it's a key factor, why dipole provides current (because of its capacitance) $\endgroup$ – user203741 Aug 22 '18 at 14:30

Dipole antenna is not just a capacitor - it also has distributed inductance and resistance - but we can still assess its charge time and discuss how this time is related to the dipole cycle.

Let's say, we applied a step voltage to the terminals of a dipole, assuming that terminals are in the middle of the dipole (center fed dipole). A positive and a negative waves will start moving outwards and, in certain time, will reach the ends of the dipole.

If we stop the clock right there (i.e., without waiting for the waves to reflect multiple times and settle), we can say that this is the charge time of the dipole capacitance and, since the velocity of these charge waves is the same as the velocity of the charges in an oscillating dipole, we can say that this time corresponds to the quarter of the dipole cycle.

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  • $\begingroup$ "A positive and a negative waves will start moving" - what the waves? You probably mean the sinusoidal(step in Your case) signal? How do they can move in space, reach the ends of the dipole, if it's just a diagram of voltage changing? $\endgroup$ – user203741 Aug 23 '18 at 16:10
  • $\begingroup$ @Artur The term wave is used here in the same sense as it is used for transmission lines or a length of rope. If we look at an individual electron in the middle of the dipole, it will move, in or out, depending on the polarity, in response to a step voltage, but it won't go all the way to the end of the dipole - it'll just push or pull (disturb) other electrons, and those electrons will do the same to their neighbors, etc., until the disturbance reaches the end of the dipole. This moving disturbance is a definition of a wave. $\endgroup$ – V.F. Aug 23 '18 at 16:40
  • $\begingroup$ Well, okay, so I was right? $\endgroup$ – user203741 Aug 23 '18 at 16:42
  • $\begingroup$ @Artur I believe so. $\endgroup$ – V.F. Aug 23 '18 at 16:44

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