Re: Slightly off topic- Van DeGraff Generator / other E&M experiments

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Original poster: "Antonio Carlos M. de Queiroz by way of Terry Fritz <twftesla-at-qwest-dot-net>" <acmq-at-compuland-dot-com.br>

Hi:

An estimation of the power required from the motor to move the belt
of a VDG generator against the electric field can be obtained:

The required electric power is simply:
P=v*i
where v is the terminal voltage and i is the belt current.
Assuming the belt operating at maximum charge density and the belt
transporting current upwards only:
i=e0*Em*A*n
where e0=8.85e-12, Em=3e6 V/m, A is the belt area (m^2), and n is the 
number of belt turns per second. Or:
i=e0*Em*W*V
where W is the belt width (m) and V is the belt speed (m/s).
The maximum terminal voltage is, for a ball terminal with radius r:
v=Em*r
And so:
P=e0*Em^2*r*W*V
Substituting the constants:
P=79.65*r*W*V

For a driving pulley with diameter d, turned by a motor turning
at rpm turns per minute:
V=rpm/60*pi*d
Resulting in:

P=4.1705*r*W*d*rpm
v=3e6*r
i=1.39e-6*W*d*rpm

Note that the power is proportional to the cube of the "size" of the
machine, the voltage proportional to the size, and the current
porportional to the square of the size, if the proportions and the 
motor speed are retained. This happens with all electrostatic machines.
Note also that the length of the belt, or the height of the terminal,
has no influence (except for insulation, and a small effect on the
output voltage, not considered).

For a typical medium machine:
r=0.25 m, W=0.1 m, d=0.05, rpm=1000
P=4.1705*0.25*0.1*0.05*1000=5.2 W
V=1000/60*pi*0.05=2.618 m/s
v=3e6*0.25=750 kV
i=1.39e-6*0.1*0.05*1000=6.95 uA
 
Double the power and the current if there is a current doubler in the 
system.
Add the mechanical load (small, but how to evaluate?) to have the 
required mechanical power.
In the example, a 100 W sewing machine motor is surely enough.

Antonio Carlos M. de Queiroz