Re: Tesla Coil RF Transmitter

["Tesla list" <tesla@xxxxxxxxxx>]

Original poster: "Bob (R.A.) Jones" <a1accounting@xxxxxxxxxxxxx>


Hi Steve and all,



> Original poster: Steve Conner <steve@xxxxxxxxxxxx>
>
>  > "In this electrostatic event energy is
>  > surely transferred
>  > without the benefit of a magnetic field or current.
>  > If you see it
>  > otherwise, please produce the experiment."
>
> This is a Gedanken experiment :P I thought about it
> some more, and I guess that physically moving a
> charged body is similar to charging and discharging a
> fixed capacitor. After all when you move the body,
> even though the charge on the body stays constant, the
> E-field seen by surrounding objects will change since
> the geometry of the system has changed.
>
> So, I hope we agree the moving body creates a
> time-varying E-field. Then, if you believe Maxwell's
> equations- the relevant one being
>
> curl H= J+dD/dt
>
> the time-varying E-field (the dD/dt term) creates a
> displacement current and this current gives rise to a
> magnetic field. Then, any mechanical work done on
> surrounding objects by "electrostatic"
> attraction/repulsion must be accounted for by E x H.

I don't believe the above is correct.
In a vacuum the dD/dt term creates a magnetic field.
The magnetic field is the same form as would be created by a current. Hence
the concept of displacement current to account for the magnetic field.

In dielectrics the field can polarize the dielectrics which is a net
separation of charges ie a real current flow which creates a magnetic field.
I think this separation of the charges is also called a displacement current
but in this case there is a real flow of charge ie a current.


> I bet that if you do all the math, the energy balance
> will add up. We have all seen the experiment where
> styrofoam packing peanuts or pie dishes get launched
> from the terminal of a HV DC power supply. If I drew a
> control surface around one of those peanuts in
> mid-flight, and calculated the integral of the
> Poynting vector over the surface, I would expect to
> find an inward EM power flow that equalled the
> mechanical power lifting and accelerating the peanut.
>
> That's how my argument goes. To be honest, I am
> starting from the biased viewpoint that Maxwell's
> equations are valid and power can't be transmitted by
> pure E or H fields- wherever you have power, there
> must be both E and H. So I invite you to put forward a
> counter argument.

If your accelerating a charge a long a line the radiation is perpendicular
to the line and proportional to the square of the acceleration.
None is emitted in the direction of the acceleration.  For example a dipole
emits no radiation end on. (in the far field: see below)

I think you will agree that if that charge approaches, head on, say a
conductive plate the voltage will change on the plate which could be used to
extract some power.
which derives from work required to move the charge.
Presumable as the charge approaches the plate the change in E field produces
some M and hence a none zero Poynting vector. Presumably this indicates that
radiation is emitted in the direction of the plate and there is no such
thing as pure E.

The above indicates we have problem. I suspect its a near field far field
problem. If the plate is in the near field it can extract power from the
near field even end on.  However the near filed falls very rapidly (faster
then squared) so there is very little power available in the near field at a
distance. i.e. dipoles don't emit end on.

I think the conclusion is we can transmitte power using the near E field
(moving charge end on) but its still involves M and its effectiveness falls
very rapidly (more than squared) with distance from the charge. The
distances are probably a few wavelengths.

We could say that in the near field we can extracting power from the E field
without the M field (but the M field is still there)  That's how  Terry's
plane wave antenna works. More accurately it would be called a near field E
detector.
Just like the flat plate E field sensors used to detect thunderstorms from
tens of  miles away.

Perhaps a more fundamental way of thinking about it is the only way charges
interact is by the exchange of photons (near field, far field, end on or
what ever on) and the Poynting vector tells you which direction the photon
is going.

>
>  > Do you really think tossing a dipole battery with
>  > positive and
>  > negative poles and equal but opposite charges out a
>  > window can create
>  > a radio wave?
>
> As long as the positive and negative poles are
> separated by a non-zero distance (ie the battery has a
> finite size) it should work as the EM effects of the
> two ends won't exactly cancel. If it spins as it
> falls, so much the better :)

Yes the above is correct. Particularly as the battery hits a hard surfaces
when the deceleration is high.
It would be possible to detect the linear motion effects close by (near
field), but at a distance (far field) only the acceleration effects.

Robert Jones