A capacitor has a plate area of 0.0045 m2 and a free charge of magnitude q on ea

Question

A capacitor has a plate area of 0.0045 m2 and a free charge of magnitude q on each plate. The space between plates has been filled with a dielectric that has a bound surface charge of magnitude 3q/4 on either side. When the dielectric material is removed, the electrical breakdown of air is observed between the plates.

Part A

What is the smallest possible value of q if the plates are isolated?

Express your answer to two significant digits and include the appropriate units.

Part B

What is the smallest possible value of q if the plates are connected to a battery?

Express your answer to two significant digits and include the appropriate units.

Explanation / Answer

part a:

let dielectric constant be k.

as we know, for a parallel plate capacitor with dielectric present,

bound surface charge=(dielectric constant-1)*free surface charge/(dielectric constant)

==>0.75*q=(k-1)*q/k

==>0.75=(1-(1/k))

==>k=4

dielectric strength of air=3*10^6 V/m

as electric field in between the plates of a parallel plate capacitor is given by E=surface charge density/epsilon

=charge/(area*epsilon)

hence for electric breakdown,

q/(0.0045*8.85*10^(-12))=3*10^6

==>q=0.0045*8.85*10^(-12)*3*10^6=1.1947*10^(-7) C

part b:

if a battery of emf V is connected to the plates,

electric field=V/d

even after removal of the dielectric , this value of electric field will remain present in between the plates.

so for dielectric break down,

V/d=3*10^6...(1)

when dielectric was present, total electric field=free charge/(epsilon*area*dielectric constant)

hence q/(8.85*10^(-6)*4*0.0045)=3*10^6

==>q=0.0045*8.85*10^(-12)*3*10^6*4=4.78*10^(-7) C

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