A parallel-plate capacitor with only air between the plates is charged by connec

Question

A parallel-plate capacitor with only air between the plates is charged by connecting it to a battery. The capacitor is then disconnected from the battery, without any of the charge leaving the plates.

A voltmeter reads 49.0 V when placed across the capacitor. When a dielectric is inserted between the plates, completely filling the space, the voltmeter reads 10.0 V . What is the dielectric constant of this material?

What will the voltmeter read if the dielectric is now pulled partway out so it fills only one-third of the space between the plates?

Explanation / Answer

when the gap is filled with air, C = A*epsilon/d

when the gap is filled with dielectric, C' = k*A*epsilon/d

here Q' = Q

C'V' = C*V

(k*A*epsilon/d)*V' = (A*epsilon/d)*V

dielctric constant, k = V/V'

= 49/10

= 4.9 <<<<<<<----------Answer

Capacitnce of the capacitor when the dielctric fills only one-third of the space between the plates

C' = (A/3)*k*epsilon/d + (2*A/3)*epsilon/d

= (A*epsilon/d)*(k/3 + 2/3)

= C*(4.9/3 + 2/3)

= 2.3*C

so, Apply Q' = Q

C'V' = C*V

2.3*C*V' = C*V

V' = V/2.3

= 50/2.3

= 21.74 volts <<<<<<<----------Answer

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