Asked by benjamin
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?
Answers
Answered by
Anonymous
i didnt get it
Answered by
Isumaina nana aishat
Answer please
Answered by
Saminu lawan
Thanks for this site
Answered by
Samip Ghimire
width (b) = 40mm = 4x10^-2 m
thickness(d)=1.5x10^-2mm=1.5X10^-5m
K=2.5
c = 2x 10^-6 F
εkA/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 F
εkA/d = 2x 10^-6 F
(εx 2.5x l x4x10^-2 )/1.5X10^-5m =2x 10^-6
l = 33.90 m
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.