Re: Displacement current
- From: "Wimpie" <wimtel@xxxxxxxxx>
- Date: 5 Sep 2006 14:27:34 -0700
Hello Chuck,
When the E-field increases from zero to the steady state value your
parallel plate capacitor is charged. The unloaded voltage between the
plates will be E*s (s=distance between plates). The capacitance of the
plates you can calculate with standard formulas. When s (distance
between plates) is very small with respect to the smallest size of the
plates, you can use: C = epsilon*Aplate/s. for vacuum epsilon =
8.854e-12 F/m, use the area of one plate for calculating the
capacitance.
Now you know the voltage and capacitance, so you can calculate the
charge on the capacitor (as function of E).
When you going to load the capacitor with a certain load you get a
transient current that will remove all charge. As probably the load
resistance will be smaller then the air resistance. The only steady
state current, after the discharge of the capacitor, will be the one
caused by the specific resistance of air, so the current you can
extract is extremely low. In case of rain drops falling on the plate
and leaving the lower plat, the current may be higher (charged rain).
When you apply an AC field (E(t) ), the easiest thing is to consider
your capacitor circuit as a voltage source with strength E(t)*s, in
series with a capacitor C. With this lumped circuit model you can
calculate the current into a certain load. This model is only valid
when the size of the plates is very small with respect to the free
space wavelength belonging to excitation frequency.
Provided that s is small with respect to plate size and Rload << Xc.
the current that will flow is caused by the time varying displacement
(d(A*E*epsilon)/dt). When you have large plates and a very sensitive
(electronic) current meter, you may be able to sense the dE/dt caused
by an approaching or leaving thunderstorm.
When your load is not floating, the situation becomes more complicated
as you get a three capacitor situation because of the ground will act
as a third plate.
For more information on the D-field, see for example Wikipedia.
Something more about the steady state current. The field line pattern
in case of AC excitation will be the same as for the DC field line
distribution. When you know the specific resistance of air, you can
calculate the resistance between the plates. The DC voltage is E*s. you
can use same equivalent circuit as used for the AC case (voltage source
with E*s with R in series).
When with an accurate formula your capacitance between the plates is
about 5% more then based on the standard parallel plate formula as
mentioned before, your R between the plates will be factor 1.05 smaller
with respect to the simple formula for calculating R (based on specific
resistance).
I hope this will help you
Wim.
.
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