On 16 Mar 2006 12:18:40 -0800, "K7ITM" wrote:

C/L = 2*pi*epsion.zero/(ln(b/a))

As b goes to infinity (the wire in freespace case), C/L goes to zero.

So if I have a thin wire, 15 meters long, out in freespace, and a 10MHz
(30 meter wavelength) EM wave comes along and passes by that wire, what
happens in the wire?

Hi Tom,

As it goes to infinity? Indeed, there are conductors and charged
bodies much closer than that to your wire in space.

However, let's look at the implications using your same formula,
except we will move L to conform to your 15 meter specification:

C = 2 · Pi · epsilon0 · L / ( ln(b/a) )
a = 1m (after all, thin is relative at infinite dimensions)
L = 15m
b= 1,000,000,000,000,000,000,000,...000m (10³³³ meters away)
epsilon0 = 0.00000000000885

C = 12 femtofarads

This was certainly at the limits of my usual Capacitor Bridge to
measure to this resolution 30 years ago, but time has marched on. This
sized capacitance is certainly encountered every day in my new field
of nanotech, and 1 femtofarad is measured by charge transfer
techniques.

Consider, Einstein's estimate of the radius of the Universe is roughly
10 Billion Light Years (±3dB) As this result above is vastly further
away than Einstein's guess (by more than 300 orders of magnitude),
lets look at again from his number:

C = 12.5 picofarads

Oddly enough, this value is on par with the distributed capacitance of
the coil's we've been pounding away on (and even more convergent, is
this is roughly the same amount of wire used in them).

73's
Richard Clark, KB7QHC