Question

A dielectric-filled parallel-plate capacitor has plate area A = 25.0 cm2 , plate separation d = 9.00 mm and dielectric constant k = 2.00. The capacitor is connected to a battery that creates a constant voltage V = 12.5 V . Throughout the problem, use ϵ0 = 8.85×10⁻¹² C2/N⋅m2 .

PART A:

U₁ was found:

U1 =

3.84×10−10

J

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.

Express your answer numerically in joules.

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.

Express your answer numerically in joules.

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?

Express your answer numerically in joules.

Answer 1

Part A.

U1 = (1/2)*C1*V1^2

C1 = k*e0*A/d

So,

U1 = k*e0*A*V1^2/(2*d)

U1 = 2*8.85*10^-12*25.0*10^-4*12.5^2/(2*9.00*10^-3)

U1 = 3.84*10^-10 J

Part B.

When dielectric is half-filled, then net capacitance of capacitor will be:

In this scenario both air filled and dielectric filled capacitor will be in parallel with each other, So

Ceq = C1 + C2

Ceq = e0*(A/2)/d + k*e0*(A/2)/d = (e0*A/(2*d))*(1 + k)

Ceq = (8.85*10^-12*25.0*10^-4/(2*9.0*10^-3))*(1 + 2.00) = 3.6875*10^-12 F

So,

U2 = (1/2)*Ceq*V^2

U2 = (1/2)*3.6875*10^-12*12.5^2

U2 = 2.88*10^-10 J

Part C.

Now when battery is disconnected, then after that charge on capacitor will remain constant, which will be equal to

Qeq = Ceq*V

Qeq = 3.6875*10^-12*12.5 = 46.09375*10^-12 C

Now when dielectric is removed then new capacitance of capacitor will be

C3 = e0*A/d = 8.85*10^-12*25.0*10^-4/(9.00*10^-3) = 2.45833*10^-12 F

Now Energy stored in capacitors is given by:

U3 = Q3^2/(2*C3) = (46.09375*10^-12)^2/(2*2.45833*10^-12)

U3 = 4.32*10^-10 J

Part D.

Work-done in removing remaining portion of dielectric will be:

W = U3 - U2

W = 4.32*10^-10 - 2.88*10^-10

W = 1.44*10^-10 J

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