A dielectric-filled parallel-plate capacitor has plate area A = 30.0 cm2 , plate

Question

A dielectric-filled parallel-plate capacitor has plate area A = 30.0 cm2 , plate separation d = 8.00 mm and dielectric constant k = 4.00. The capacitor is connected to a battery that creates a constant voltage V = 7.50 V . Throughout the problem, use 0 = 8.85×1012 C2/Nm2 .

Part A

Find the energy U1 of the dielectric-filled capacitor.

Part B

The dielectric plate is now slowly pulled out of the capacitor, which remains connected to the battery. Find the energy U2 of the capacitor at the moment when the capacitor is half-filled with the dielectric.

Part C

The capacitor is now disconnected from the battery, and the dielectric plate is slowly removed the rest of the way out of the capacitor. Find the new energy of the capacitor, U3.

Part D

In the process of removing the remaining portion of the dielectric from the disconnected capacitor, how much work W is done by the external agent acting on the dielectric? Sorry this is a long one but I would really appreciate it.

Explanation / Answer

a)

Capacitance

Cd=KeoA/d =4*(8.8542*10-12)*(30*10-4)/(8*10-3)

Cd=1.33*10-11F

Energy stored in the capacitor

U1 = (1/2)C1V2=(1/2)(1.33*10-11)*7.52

U1=3.74*10-10J

b)

Now equivalent capacitance is parallel combination of capacitance with dielectric and without dielectric

Ceq=C1+C2 = Keo(A/2)/d +eo(A/2)/d =(K+1)eoA/2d

Ceq = (4+1)*(8.8542*10-12)(30*10-4)/2*(8*10-3)

Ceq=8.3*10-12 F

energy stored

U2=(1/2)CeqV2 =(1/2)*(8.3*10-12)*7.52

U2=2.33*10-10 J

c)

Now Charge remains constant

U2=(1/2)CeqV2 =(1/2)QV

since Q=CV

Q=2U2/V =2*(2.33*10-10)/7.5 =6.23*10-11 C

Capacitance without dielectric

C=Cd/K =1.33*10-11/4=3.32*10-12F

Energy stored

U3=(1/2)(Q2/C)=(1/2)(6.23*10-11)2/(3.32*10-12)=5.84*10-10 J

d)

Work done

W=dU=U3-U2 = 3.5*10-10 J

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