A parallel-plate capacitor has plates of area 2.90x10-4 In Part A What plate sep

Question

A parallel-plate capacitor has plates of area 2.90x10-4 In Part A What plate separation is required if the capacitance is to be 1420 pF ? Assume that the space between the plates is filled with air? (Dielectric constant for air is 1.00059) Express your answer using three significant figures. d1.809 10 SubmitP XIncorrect; Try Again; 3 attempts remaining Part B What plate separation is required if the capacitance is to be 1420 pF ? Assume that the space between the plates is filled with paper. Dielectric constant for paper is 3.7) Express your answer using two significant figures Submit Request Answer

Explanation / Answer

Capacitance, C = 1420 pF = 1420*10-12 F

Plate area, A = 2.90*10-4 m2

Eqn for the Capacitance of a parallel plate capacitor is given by

C = 0rA/d

or

d = 0rA/C

Part A:

Separation between the plates is filled with Air:

here dielectric constant, r = 1.00059

So plugging in the values,

d = 0rA/C

d = [8.85*10-12*1.00059*2.90*10-4 m2] / [1420*10-12 F]

d = 1.81* 10-6 m

d = 1.81 um

Part B:

Separation between the plates is filled with Paper:

here dielectric constant, r = 3.7

So plugging in the values,

d = 0rA/C

d = [8.85*10-12*3.7*2.90*10-4 m2] / [1420*10-12 F]

d = 6.691* 10-6 m

d = 6.691 um

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