1) A large capacitance of 1.01 mF is needed for a certain application. (a) Calcu
ID: 1397566 • Letter: 1
Question
1) A large capacitance of 1.01 mF is needed for a certain application.
(a) Calculate the area the parallel plates of such a capacitor must have if they are separated by 4.11 µm of Teflon, which has a dielectric constant of 2.1.
________ m2
(b) What is the maximum voltage that can be applied if the dielectric strength for Teflon is
60 106 V/m
______________ V
(c) Find the maximum charge that can be stored.
_______________C
(d) Calculate the volume of Teflon alone in the capacitor.
_____________m3
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Find the total capacitance of the combination of capacitors shown in the figure below.
(C1 = 2.52 µF, C2 = 21.7 µF.)
µF
Explanation / Answer
capacitance of the capacitor = 1.01 mF or 1.01 * 10^-3 F
dielectric strength of the teflon = 2.1
capacitor is seperated by = 4.11 * 10^-6 m
capacitance = e * A / d
1.01 * 10^-3 = 2.11 * A / 4.11 * 10^-6
A = 1.97 * 10^-9 m^2
the area the parallel plates = 1.97 * 10^-9 m^2
maximum voltage = dielectric strength * thickness of dielectric
maximum voltage = 60 * 10^6 * 4.11 * 10^-6
maximum voltage = 246.6 V
charge = capacitance * voltage
charge = 1.01 * 10^-3 * 246.6
maximum charge = 0.249066 C
volume = area * thickness of the dielectric
volume = 1.97 * 10^-9 * 4.11 * 10^-6
volume of the dielectric = 8.1 * 10^-15 m^3
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