Re: Tesla Coil RF Transmitter

["Tesla list" <tesla@xxxxxxxxxx>]

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 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.



> 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 :)

Steve
http://www.scopeboy.com/