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Determining Capacitance?
Determine the capacitance of a Teflon-filled parallel-plate capacitor having a plate area of 1.85 cm2 and insulation thickness of 0.0400 mm.
I tried using the formula, C = ε₀ ∙ A / d:
Giving me C = (8.854e-12)(1.85e-4)/(0.04e-3) = 4.09e-11 * (1e+12) = 40.9 pF.
However the software tells me this answer is wrong. The answer is somewhere in the range of 10~100pF, but I don't want to randomly guess at it. Could someone explain to me how to solve this problem?
The answer is 86 pF. Could someone explain how they got that answer?
2 Answers
- 1 decade agoFavorite Answer
You've not accounted for Teflon. Probably because your formula is weird. It's C=(Kε) (a/d) (8.85e-12)
Kε being the dielectric constant, and in this case it is Teflon which has a dielectric constant of 2.1
- deloyLv 45 years ago
The voltage on a capacitor charging, is given via Vc = V [one million - e^ ( -t / R x C) ] 9 = 13 [ one million - e^ ( -one million / 11E3 x C ) ] 9 = 13 - 13 e^ ( -one million / 11E3 x C ) ] 13 - 9 / 13 = e^ ( -one million / 11kE3 x C ) ] ln ( 4 / 13 ) = -one million / 11E3 x C -one million.179 = -one million / 11E3 x C one million /one million.179 = 11E3 x C 0.848 = 11E3 x C C = 0.848 / 11E3 = 77 E-6 F = 77 microfarads .