At 7:22 PM 4/5/5, John Berry wrote:
>Great analysis.
>It should work IMO even if the forces are as Ampere states unless there
>is no displacement current,
[snip]
There is always a displacement current, even between vacuum plates. The
displacement current is eactly equal to the current to the capacitor,
regardless of the presence of a (non-vacuum) dielectric or not. A little
proof follows that the H generated by the changing E of a capacitor is
identical to that from a conductor current, i.e the displacement current is
equal to the capacitor current.
Here is the basic picture:
xxxxxxxxxxxxxx
x x
x x P1
x -----------
power gap <==== hoped for thrust
x -----------
x x P2
x x
xxxxxxxxxxxxxx
Note: here interpret "@" as the partial derivitive symbol below.
The capacitance of the capacitor is:
C = epsilon A/d
where A is the plate area and d is the separation. The conduction current
is then:
i_c = C dv/dt = (epsilon A/d) dv/dt
On the other hand, the electric field in the dielectric, be it pure vacuum
or not, is, neglecting fringing (which doesn't occur significantly in the
coaxial version anyway):
E = v/d
Hence:
D = epsilon E = (epsilon/d) v
@D/@t = (epsilon/d) v/dt
and i_d, the displacement current, is (D normal to the plates) given by
(now using D as a current density vector, S a surface envelope):
i_d = integral over A{ @D/@t dot dS }
= integral over A{ (epsilon/d) dv/dt dS}
= integral over A{ (epsilon A/d) dv/dt }
= i_c
so i_d, the displacement current, is always equal to the capacitor current ic.
Regards,
Horace Heffner
