Michael Foster wrote:
I'm trying to figure out how a couple of guys who are clearly
better educated, and probably a lot smarter than I, can have
gone so far wrong.
OK it's time to point out something trivial.
divergence(E) = 4 pi rho (in cgs units)
E is the "real" electric field (not the "auxiliary" field) and "rho" is
the actual charge density (including both mobile and fixed charges in
conductors, insulators, dielectrics, and what have you).
That means that for any spherically symmetric distribution of charge,
with center at O, the field at any point P depends _ONLY_ on the total
quantity of charge closer to O than point P. That's all. You can use
any insulator you like, and if it's also spherically symmetric, it won't
affect the field outside ... unless it carries a net charge.
Maxwell's equations are generally well tested in the realm of
electrostatics.
I'm afraid that, throughout the following, there's been some confusion
between voltage and field strength. "Lift" -- force on the apparatus --
has to do with field strength, not voltage. Details follow.
Stephen A. Lawrence wrote:
Frederick Sparber wrote:
Posted earlier:
This Field Line Applet is cheaper than buying more VDGs.
http://www.cco.caltech.edu/~phys1/java/phys1/EField/EField.html
If the force around a positively charged sphere is reduced by
cladding it with a high dielectric constant material (K)
It won't be, not unless the cladding has a net charge. The
dielectric or other properties of the cladding aren't relevant
to its ability to "shield" something, which can be done only
by canceling the original field. (There's no such thing as a
"true" electric shield, of course. At least according to
conventional theory, the E field obeys the law of
superposition,and you can't actually "block" it; you can
only cancel it.)
If you surround a charged sphere with any spherically
symmetric material which does not, itself, carry a net charge,
the result will be no change in the external field (outside
the cladding) and no change in the net force acting on the
system (sphere+cladding) due to an external field.
This is demonstrably not the case. If a polar dielectric is
interposed between a charged conductor and a detector,
What, exactly, is the detector detecting? Field strength, or _voltage_?
(And what /shape/ is the dielectric? Large flat sheet, or something
else?)
Note well: Voltage and electric field strength are different properties.
The voltage is the path integral of the electric field; the electric
field is the gradient of the voltage; the field strength is the
magnitude of the electric field.
Inside a polar dielectric in a strong field there is a reverse field due
to the dipoles in the material, which _certainly_ affects the voltage
along a path which runs through the dielectric. If you drop a charged
particle "down" the field to the dielectric, through a tiny hole in the
dielectric, and on its way, it will speed up until it gets to the
dielectric, slow down during its passage through the tunnel, and then
start speeding up again on the other side. The _voltage_ is the net
energy gain of the particle over the whole path, and during its
traversal of the dielectric it loses energy, due to the reverse field
inside. Hence, the voltage is lower with the dielectric in place.
The _field strength_ is how fast the particle accelerates at any
particular point on the path. In other words, field strength equates to
force -- and "lift" is upward-directed "force".
the
strength of the field will drop substantially,
Only _inside_ the dielectric!
especially if
the dielectric has a high K. Work is done by aligning the
molecular dipoles with the result that the voltage drops while
the coulombs increase. When the dielectric is removed the
voltage increases again.
You've said two different things here: "the strength of the field will
drop", and "the voltage drops". The dielectric will _certainly_ affect
the voltage, just as interposing a charged parallel plate capacitor
would affect the voltage (which would reduce it by the voltage to which
the capacitor had been charged). As an aside, that is how a polar
dielectric acts to increase the capacity of a parallel plate capacitor
versus using an air dielectric: for any particular net charge on the
plates, interposing it reduces the voltage between the plates;
consequently the ratio of charge to voltage increases.
What it won't do, however, _if_ it's a large flat sheet and the field is
perpendicular to it (and it carries no net charge), is have a
significant effect on the electric field on either side of it ... any
more than a large parallel plate capacitor would.
If you disagree please explain either (a) how a polar dielectric in an E
field differs in its field from a charged capacitor or (b) how either
one of them has a large effect on the field outside the "plates".
As a practical matter, you can feel the strong attraction
between the polar dielectric and the charged conductor,
It's got a dipole field, even though its monopole field in the area near
the middle of either surface is insignificant. The thicker it is, the
stronger the dipole field.
and
as the dielectric approaches the charged object the attractive
force between the charged conductor and other objects which
may have been attracted initially is noticeably diminished.
If you have an electrostatic voltmeter, the voltage drops
as the dielectric approaches the charged conductor.
As I already said, it affects the voltage. That was never at issue.
This effect is not observed if the interposed insulator is a
non-polar material, polyethylene for example. For a really
unequivocal demonstration, you want a material with K > 100.
Glass works, but its dielectric constant is only about 7.
You'd like to use barium titanate, but a free-standing plate
of this stuff is almost impossible to come by. I like to use
the T.T. Brown standby, litharge in epoxy, K = approx 700.
If it did you could close the cladding, move a bit, open the
cladding, move back, close the cladding, and so forth, and
voila, you have a PMM. With uncharged cladding opening and
closing the cladding (by sliding it) should require
insignificant work.
Well, no. The only time lateral transposition of the
dielectric would require no work would be if the charged
object was standing alone in a universe of no other charged
or conductive objects.
You're right; it would be pulled across the gap; I overlooked that.
But the force would be independent of the motion, so you could, in
principle, regain the energy you expend to open the shield when it's
closed again, for re-use at the next opening; clever use of springs
might do the job.
One of the old classic electrostatic
demonstrations is the drawing in of a dielectric sheet
between the plates of a charged air gap capacitor. Work in,
work out.
If it's "drawn in", then it's energy gain in, work out.
In Fred's proposed setup, however, as there would be a
continous supply of current provided by the Van de Graaff
generators, the voltage at the surface of the dielectric
covered sphere would eventually be the same as a the three
outlying bare metal spheres. It would just take longer to
get there.
A better solution would be to expel the positive charge at
the juncture of the three Van de Graaff generators with a
cluster of needle points, perhaps assisting the the hoped-
for lift with an ion wind.
Which would be the /only/ lift, if the overall apparatus has no net charge.
Fred wrote:
Force = 1/K * 1/4(pi)eo * q* Q/R^2
Hence a 3-point craft with a cladded positive center sphere
and three exposed negative spheres should repel the earth's
excess negative charge up to an altitude that requires charge
reversal for getting past the ionosphere.
There is a fundamental problem with this idea. While the earth
has a net negative charge of say, one megajoule, the tiny
fraction of a joule per square meter just won't supply the
repulsive force you need unless your Van de Graaff spacecraft
is very large and already elevated. The earth is is quite
conductive, especially at the voltages we are talking about here.
Therefore, an object with a strong negative charge will charge
the earth locally positive by induction and your spheres will
be strongly attracted to the earth, not repelled. The earth's
negative field will have been slightly shifted to the opposite
side of the world.
Yes, for sure -- the Earth, on this scale and with static charges, can
be viewed as a conductor, and in response to a charge held near it, it
produces the effect of an "image charge" of opposite polarity
equidistant from the surface of the Earth (but underground).
If you want to play around with a propulsive force, I suggest
you try some Brown-Biefeld experiments. I've had numerous
positive results, but haven't written about them because I'm
not sure I've eliminated all the artifacts such as ion wind.
Charge on.
M.
