How does the high voltage get to the ends?

In short, the electromagnetic field set up around the antenna tend to shove a high density of electrons to one end or the other at different parts of the RF cycle. 

In the middle of the antenna, the electrons can easily travel further down the antenna, but when they reach the antenna ends and the electric field starts getting strong, they pile up there. 

As far as the DANGER goes, something you need to know is that RF tends to travel on the outside of your body when you come in contact with it.  The conductivity of your body tends to exclude currents from traveling straight through you.

This is called skin effect, but it's named for the effect in other conductors, not the effect on people :-)  Electromagnetic fields can't penetrate far into a good conductor.  A perfect conductor would exclude them entirely.  So the current density in a conductor decays exponentially as you go in from the outside.

This means that bridging hundreds of volts of RF is much less serious (i.e. not fatal), where bridging several hundred volts of 60Hz or DC can kill you dead, while the same RF voltage will probably just hurt and maybe give you a nasty black spot where you get a "RF burn."  I read some cases of people who came into contact with quite high voltage RF and got a serious full body RF burn with some lingering health effects... but there don't seem to be many cases of fatal RF electrocution even from very high power transmitters.  You don't want to mess around but it's not like you're playing with 1kVDC...

Is the danger as great at the center of a dipole?

Nope, not even close.  I know it's weird for the voltage between the two dipole halves to depend on where you are on the dipole, but it's because of the way the electromagnetic field is set up around the antenna.  Near the middle, there's a lot of magnetic field encircling the wire.  At the ends, there's a strong electric field that "starts" on one end and goes to the other, if you like thinking about field lines.

Of course, at EVERY point on the dipole there's some electric field ending on the wire and some magnetic field encircling it, but at the middle, there's a lot more magnetic field from a high current flowing in the wire, and much weaker electric field.  So the voltage from one dipole half to the other at the feedpoint is low compared to the ends... a hundred volts peak (70.7V RMS) at 100W into a 50 ohm load.   

It is true that coming into contact with antenna tips even at low power isn't nice.  I got my first RF burn when I was a ham... running 4W of CB power into a half square I built.  I touched the end of that and it really stung.

The nastiest RF burn I got was off a 10m matching network on a random wire I was adjusting with 30-40W or so.  That sucker made a tiny hard black scorched spot on my skin  that lasted for several weeks.  The problem with high voltage RF or lower frequency, of course, is that it easily ARCS to you as you pull away.  And at high power arcing to trees or whatever can pretty easily start fires.

So don't mess around, but also don't think that you can casually kill yourself with a low power transmitter's output and the voltage step up inherent in a dipole antenna.  It's very high voltage, but it's not like working inside an amplifier or contacting high voltage power lines, just because the current can't get inside your body easily.

73
Dan

There are several equally good ways to look at this.

The ends have nothing terminating them. At the open ends, antenna current has nothing to flow into except a very small amount of capacitance. Since the antenna is also a single wire transmission line, the surge impedance of the antenna transforms the very high impedance at the open ends down to a low impedance 1/4 wave away from the ends. In other words the antenna acts like an open ended single wire transmission line.

Another more esoteric way to look at it is with standing waves. An open circuit inverts current 180 degrees. An open circuit does not invert voltage. Because current is inverted 180 degrees, the reflection from the open end "cancels" the current. Because the voltage is not inverted at the open ends, the voltage goes very high. It is only limited by the surge impedance of the single wire trnamssion line making up the antenna and the capacitance and electric field boundary at the open end. 1/4 wave away the voltage is out of phase, and the current is in phase. This makes current high 1/4 wave in from the open end.

It is this voltage distribution that sets up the induction fields around the antenna. The electric field is very high at the ends because of the standing waves on the antenna, and because there is a very sharp boundary between the antenna end and space around the antenna. The magnetic field is stronger near high current areas. Radiation, a totally different effect, is caused by the charges accelerating and so is more intense near the high current areas of the antenna where there are many more charges "wiggling".

The fields are actually created by what is going on in the wire, not the other way around.

The most simple concept is that the open circuit ends cannot support much current because the impedance is very high, and for a given power that means voltage is very high compared to current. Since the "load" is the open end of the antenna, and since it acts like a transmission line, the impedance is progressively lower until 1/4 wave back from the open end.

If it is a closed loop like a dipole, you simply look at what happens exactly opposite the feedpoint. Exactly opposite the feedpoint is a short, and the short between the ends means current is maximum. 1/4 wave back from that short is a voltage maximum and current minimum, again because of transmission line effects. 1/4 wave more and the high current area maximum repeats.

73 Tom

Thanks to all of you! I think that I now understand what I was asking about. This is almost as good as having a bunch of real life elmers to call on!.
Don

Hi Tom:

  Actually, I think the the concept of a single wire transmission line is more esoteric than your latter example.  In fact, I'm not convinced a single wire transmission line can exist in free space.  However, over a ground surface...of course it can.  I actually have a TRUE Windom that relies on this...probably one of a handful of hams still breathing who actually knows what a true Windom is!

   Fortunately, all the analogies presented in this thread are useful in explaining this voltage distribuition in one way or another.  Even more significantly (and curiously), the concept also applies to DC!  If you take a Van DeGraff generator for example, the highest voltage will exist on the SURFACE of the sphere...whether it's a hollow ball or a solid sphere of brass!  This is because the charges all repel, and are all forced to the surface.  When you reach the "end of the road" there's nowhere to go, so the charge HAS to accumulate there.

eric

A single wire transmission line can certainly exist is freespace. It has an electric field meaning it has capacitance distributed along its length, and it has series inductance.

Some microwave antennas launch a wave along a single wire transmission line. No return path required. Just a cone at each end.

G line exploits a surface wave on the wire, and requires an excitation that excites that mode and not the one that couples well to free space waves.

73
Dan

That would require that the wave is perpendicular to the surface of the wire....in other words, vertically polarized in all directions.  

I guess the conical launchers sort of do that.

Voltage with respect to the other end of the dipole is useful too!

However, most of us folks who've worked with high power know that you can have corona off the end of a dipole or other antenna with NO nearby "reference."  This  generally means you have a surplus or deficit of electrons relative to the surrounding air.

Eric