± Energy of a Capacitor in the Presence of a Dielectric A dielectric-filled para

Question

± Energy of a Capacitor in the Presence of a Dielectric

A dielectric-filled parallel-plate capacitor has plate area A = 15.0 cm2 , plate separation d = 10.0 mm and dielectric constant k = 2.00. 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.

Express your answer numerically in joules.

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.

Explanation / Answer

A)
C= Ebsolene * A/ d
Here
Ebsolene = k * Ebsolene0
= 2 * (8.85*10^-12)
= 1.77 *10^-11
A= 15 cm^2 = 1.5*10^-3 m^2
d = 10mm = 0.01 m

C= Ebsolene * A/ d
= (1.77 *10^-11 ) * (1.5*10^-3) / (0.01)
= 2.655*10^-12 F

V= 7.5 V

Energy, U1 = 0.5 * C*v^2
= 0.5 * (2.655*10^-12 ) * 7.5^2
= 7.5*10^-11 J

Part B:
When dielectric plate is half pulled out:
C (without dielectric)= Ebsolene0 * A/ (d/2)
= (8.85 *10^-12 ) * (1.5*10^-3) / (0.01/2)
= 2.655*10^-12 F

C(with dielectric)= Ebsolene * A/ (d/2)
= (1.77 *10^-11 ) * (1.5*10^-3) / (0.01/2)
= 5.32*10^-12 F

This is like both these capacitors are connected in series,
so Cnet = C1C2/ (C1+C2)
= (2.655*10^-12 ) * (5.32*10^-12 ) / (2.655*10^-12 + 5.32*10^-12 )
= 1.41*10^-23 / (7.975*10^-12)
= 1.768*10^-12 F

V= 7.5 V

Energy, U2 = 0.5 * C*v^2
= 0.5 * ( 1.768*10^-12 ) * 7.5^2
= 4.97*10^-11 J

PArt C:
Charge on tthe capacitor when dielectric is half inside (above scenario)
Q= C* V
= 1.768*10^-12 * 7.5
= 1.326*10^-11 C

This charge will remain constant when battery is completely pulled out
After dielectric is completely pulled out:
C = Ebsolene0 * A/ d
= (8.85 *10^-12 ) * (1.5*10^-3) / (0.01)
= 1.3275*10^-12 F

Energy, U3 = 0.5 Q^2/C
= 0.5 * (1.326*10^-11 )^2 / ( 1.3275*10^-12)
= 6.62*10^-11 J

Part D:
work done, W = increase in energy
= U3 - U2
= 6.62*10^-11 - 4.97*10^-11
= 1.65*10^-11 J

Related Questions

Navigate

• Browse (All)
• Browse #
• Subjects

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