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If we know the attraction force and the capacitance, then we can calculate
the voltage. If the attraction force is 0.1Nt, and is constant
with distance because the plates are closely spaced, and if the spacing of
the "capacitor plates" is initially 1mm (0.001meter), then it took F*d
Joules to pull the attractive plates apart to 1mm, and the stored energy
is .1Nt * .001meter = 1e-4 joules. OK so far? Now, what is the
capacitance of two plates spaced 1mm apart and having the size of the
contact area between the balloon and your forearm? Lets say the area is
4cm by 15cm, or .04*.15 = .006 square meters. The capacitance equation is
C = k*area/spacing, where k=8.9e-12 and lengths are in meters, so the
capacitance is 53.4pf. The energy stored was 100 microjoules, so the
voltage can be had by E=1/2CV^2.
I get an answer of 1.9 kilovolts. That's at 1mm spacing. The .1 Nt
attraction sets the charge on each plate at a constant .1 microcoulombs:
Fd=1/2Q^2/C, sqrt(2CFd) = 1e-7coulombs. Capacitor voltage is then V =
Q/C, and capacitance varies inversely with plate-spacing if the
fringing fields are ignored. Therefor voltage between the plates varies
directly with plate spacing:
plate spacing V(capacitor)
1mm 1917v
5mm 9584v V = (1e-7)*D/(.006)/(8.9e-12)
1cm 19168v
5cm 95840v
Rubbing a balloon on arm-hair creates tens of kilovolts! The voltage
at the greater spacings would be too high, because the gap between
plates is approaching the size narrow dimension of the plates, and
fringe fields become important.