12. O-10 points SerCP11 16 P.055 My Notes Ask Your Teacher (a) Determine the cap
ID: 1580710 • Letter: 1
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12. O-10 points SerCP11 16 P.055 My Notes Ask Your Teacher (a) Determine the capacitance that can be applied to a Teflon-filled parallel-plate capacitor having a plate area of 235 cm2 and insulation thickness of 0.0620 mm nF (b) Determine the maximum voltage that can be applied to a Teflon-filled parallel-plate capacitor having a plate area of 235 cm2 and insulation thickness of 0.0620 mm kv Need Help? Read t 13. -120 points OSColPhys1 19.5.064 XP My Notes Ask Your Teacher A large capacitance of 2.00 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 5.00 um of Teflon. b) What is the maximum voltage that can be applied? (c) Find the maximum charge that can be stored. (d) Calculate the volume of Teflon alone in the capacitor. m2 m2 Additional MaterialsExplanation / Answer
a) capacitance C = A*Eo/d
=> C = 235*10^-4*8.854*10^-12/(0.062*10^-3) = 54.12 nF
b) breakdown voltage of Teflon insulating film = 173*10^6 v/m
=> max voltage that can be applied ,V = 173*10^6*0.062/1000 = 10.726 kV
c) C = AEo/d
=> A = C*d/Eo = 2*10^-3*5*10^-6/(8.854*10^-12) = 1129.43 m^2
d) max voltage that can be applied , V = 173*10^6*5*10^-6 = 865 V
e) max charge , Q = CV = 2*10^-3*865 = 1.73 C
f) volume of teflon = A*separation = 1129.43*5*10^-6 = 0.005647 m^3
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