A dielectric-filled parallel-plate capacitor has plate area A = 20.0 cm2 , plate

Question

A dielectric-filled parallel-plate capacitor has plate area A = 20.0 cm2 , plate separation d = 7.00 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 .

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.

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.

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?

Explanation / Answer

A) Capacitance = k.0.A/d
= 2.00 * (8.85 * 10^-12) * (20.0 * 10^-4) / (7.00 * 10^-3)
= 5.06 * 10^-12 F

Energy (U1) stored in that capacitor when charged to 7.5 V:
U1 = ½CV²
= ½ * 5.06 * 10^-12 * 7.5²
= 1.42 * 10^-10 J

B) The dielectric is half removed from between the plates, with the battery still attached, so that effectively, you have two capacitors in parallel, each with half the plate area, one with dielectric of k = 2.00 and one with air (k = 1)

Ceq = 2.00 * (8.85 * 10^-12) * (10 * 10^-4) / (7.00 * 10^-3) + (8.85 * 10^-12) * (10 * 10^-4) / (7.00 * 10^-3)
= 3.79 * 10^-12 F

Giving us a new stored energy:
U2 = ½CV² = ½ * 3.79 * 10^-12 * 7.5² = 1.066 * 10^-10 J

(C) At this point the battery is disconnected, so the charge (Q) on the plates is now trapped. {Before you start to move the dielectric any further, the voltage between the plates remains at 7.5 volts}

Q = CV = 3.79 * 10^-12 F * 7.5 V = 2.84 * 10^-11 coulombs

Pull out the dielectric. What is the new capacitance with k = 1?
capacitance = k.0.A/d
= 1.00 * (8.85 * 10^-12) * (20.0 * 10^-4) / (7.00 * 10^-3) = 2.53 * 10^-12 F

Charge is fixed, capacitance has changed, so the voltage between the plates must change.
Q = CV gives
V = Q/C = (2.84 * 10^-11) / (2.53 * 10^-12) = 11.23 V
U3 = ½CV² = ½ * 2.53 * 10^-12 * 11.23² = 1.59 * 10^-10 J

D) Recall that U2 was 1.066 * 10^-10 J
So the extra energy (U3 - U2) = 5.24 * 10^-11 joules
Work done by external agent = 5.24 * 10^-11 joules

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