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

Question

A dielectric-filled parallel-plate capacitor has plate area = 30.0 , plate separation = 10.0 and dielectric constant = 4.00. The capacitor is connected to a battery that creates a constant voltage = 15.0 . Throughout the problem, use = 8.85×10-12 .
A) Find the energy of the dielectric-filled capacitor
B) The dielectric plate is now slowly pulled out of the capacitor, which remains connected to the battery. Find the energy 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.
D) In the process of removing the remaining portion of the dielectric from the disconnected capacitor, how much work is done by the external agent acting on the dielectric?

Explanation / Answer

C = e*A/d, where e = 3* (8.85x 10^-12) C = 443 pF 1) U1= 1/2(443pF)(15^2) = 5 x 10^-8 joules 2) with the dielectric half in the capacitor C is equal to 71.67 + 221.5 = 228.67 pf U2 = 1/2(228.78 pF)(15^2) = 2.57 x 10^ -8 joules Q = CV = (228.67 pF)(15 V) = 3.43 x 10^ - 9 coulombs 3) C = 148 pF Q = 3.43 x 10^ -9 coulombs V = Q/C = 3.43 x 10^ -9/1.48 x 10 ^ -10 = 23.17 volts U3 = 1/2(148 pF)(23.17)^2 = 3.97 x 10^-8 joules 4) 0.54 x 10^ -8 joules is the work done pulling out the dielectric. This make U3 = U2 + 0.54 x 10^ - 8 joules

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