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 50.0 V when placed across the capacitor. When a dielectric is inserted between the plates, completely filling the space, the voltmeter reads 15.0V 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'

= 50/15

= 3.333 <<<<<<<----------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*(3.33/3 + 2/3)

= 1.78*C

so, Apply Q' = Q

C'V' = C*V

1.78*C*V' = C*V

V' = V/1.78

= 50/1.78

= 28.1 volts <<<<<<<----------Answer

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.