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 47.0 V when placed across the capacitor. When a dielectric is inserted between the plates, completely filling the space, the voltmeter reads 14.0 V . What is the dielectric constant of this material?

K=??

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?

V=??

Explanation / Answer

Charge will be same in both condition

hence Q1 = Q2

C1V1 = C2V2

e0A/d*V1 =e0KA/d*V2

K = V1/V2 =47/14 =3.357

(b) Ctotal =Ca + Cb

Ca = K*e0A/3d

Cb =2*e0A/3d

Ctotal =e0A/d(K/3 + 2/3)

V = C1V1 / Ctotal = V1 / (K/3 + 2/3) =47/(3.357/3 + 2/3) =26.32 V...........Ans.

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