PARTS A THROUGH D Consider a parallel-plate capacitor made up of two conducting

Question

PARTS A THROUGH D

Consider a parallel-plate capacitor made up of two conducting plates with dimensions 29 mm times 32 mm. If the separation between the plates is 1.8 mm, what is the capacitance, in pF, between them? If there is 0.93 nC of charged stored on the positive plate, what is the potential, in volts, across the capacitor? What is the magnitude of the electric field, in newtons per coulomb, inside this capacitor? If the separation between the plates doubles, what will the electric field be if the charge is kept constant?

Explanation / Answer

part(a)

capacitance C = e0*A/d

A = area of the plates = 0.029*0.032 m^2

d = sepeartion between plates = 1.8*10^-3 m

C = 8.85*10^-12*0.029*0.032/(1.8*10^-3)

C = 4.56*10^-12 F <<<<------------ANSWER

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part(b)

potential V = Q/C = 0.93*10^-9/(4.56*10^-12) = 203.9 V

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part (C)

Electric field E = V/d = 203.9/(1.8*10^-3) = 113277.8 N/C

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part (D)

Eelctric field E = Q/(A*e0)

If the seperation between plates doubles

the electric field remains same <<<----------------ANSWER

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