WA Hw 7 BP 115 HYS Chegg C www.webassign.ne 13509770 t/web/Student/Assignment-Re

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WA Hw 7 BP 115 HYS Chegg C www.webassign.ne 13509770 t/web/Student/Assignment-Responses/submit?dep 2 points. Previous Answers Walker4 20.P.057 My Notes 4 m2. What plate separation is required if A parallel-plate capacitor with plates of area 2.90 x 10 when the space between the plates is fille with the capacitance is to be 1330 pF, each of the following? air 1.93e-6 (b) rubber 460902.256 Your response differs significantly from the corre answer. Rework your solution from the beginning and check each step carefully m 3/3 points Previous Answers Walker4 20.P.065 My Notes can be approximated by a parallel-plate capacitor with plates of area 5.85 x 10 m2, a plate separation of 8.5 x 10 9 m, and a dielectric with a The membrane of a living cell dielectric constant of 4.5. a) What is the energy stored in such a cell membrane if the potential difference across it is 0.0633 V 5.50e-14. b) Would your answer to part (a) increase, decrease, or stay the same if the thickness of the cell membrane is decreased increase decrease stay the same Explain Because if thickness decreased, capacitance would increase 6:52 PM 3/8/2016

Explanation / Answer

here,

area of plates , a= 2.9 * 10^-4 m^2

capacitance , C= 1330 * 10^-12 F

dielectric constant for rubber , k = 2

let the sepration between them be d

C = K * A * e0 /d

1330 * 10^-12 = 2 * 2 * 10^-4 * 8.85 * 10^-12 /d

d = 2.7 * 10^-6 m

the sepration between plates is 2.7 * 10^-6 m

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