A sheet of paper 40mm wide and (1.5 x 10^-2) thick is placed between a metal foil of the same width is used to make a 2.0 UF capacitor. If the dielctric constant(relative permitivity of the paper is 2.5 ,what length of paper is required?
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width (b) = 40mm = 4x10^-2 m
thickness(d)=1.5x10^-2mm=1.5X10^-5m
K=2.5
c = 2x 10^-6 A/d = 2x 10^-6 F
(εx 2.5x l x4x10^-2 )/1.5X10^-5m =2x 10^-6
l = 33.90 m
thickness(d)=1.5x10^-2mm=1.5X10^-5m
K=2.5
c = 2x 10^-6 A/d = 2x 10^-6 F
(εx 2.5x l x4x10^-2 )/1.5X10^-5m =2x 10^-6
l = 33.90 m
3 answers