± Energy of a Capacitor in the Presence of a Dielectric Part A Find the energy U

Question

± Energy of a Capacitor in the Presence of a Dielectric Part A Find the energy U1 of the dielectric-liled capacitor Express your answer numerically in joules. A dielectric-filled parallel-plate capacitor has plate area A 30.0 cm2. plate separation d 6.00 mm and dielectric constant k 5.00. The capacitor s connected to a battery that creates a constant voltage V- 12.5V. Throughout the problem, use 85x10-12C"/N·m? U1 Submit Hints My Answers Give Up Review Part Part B The dielectric plate is now slowly pulled out of the capacitor, which remains capacitor at the moment when the capacitor is hail-fSled with the dielectric Express your answer numerically in joules connected to the batery Find the energy Uh of the Us 4166

Explanation / Answer

epsilon_0 = 8.854 *10^-12

k = 5

d = 6 mm = 0.006 m

area , A = 30 *10^-4 m^2

V = 12.5 V

part A)

U1 = 0.5 * C * V^2

U1 = 0.5 * k * epsilon_0 * area * V^2/d

U1 = 0.5 * 5 * 8.854 *10^-12 * 30 *10^-4 * 12.5^2/.006

U1 = 1.729 *10^-9 J

the energy stored in dielectric capacitor is 1.729 *10^-9 J

part B)

when the capacitor is half filled with dielectric

U2 = 0.5 * C * V^2

U2 = 0.5 * (k +1) * epsilon_0 * area/2 * V^2/d

U2 = 0.5 * (1 + 5) * 8.854 *10^-12 * 30 *10^-4 * 12.5^2/(2 *.006)

U2 = 1.038 *10^-9 J

part C)

charge on the capacitor , Q = (k +1) * epsilon_0 * area/2 * V/d

Q = (1 + 5) * 8.854 *10^-12 * 30 *10^-4 * 12.5/(2 *.006)

Q = 1.66 *10^-10 C

capacitance , C = 8.854 *10^-12 * 30 *10^-4/(.006)

C = 4.427 *10^-12 F

energy stored = 0.5 * Q^2/C

energy stored = 0.5 * (1.66 *10^-10)^2/(4.427 *10^-12)

energy stored = 3.11 *10^-9 J

the energy stored , U3 is 3.11 *10^-9 J

part D)

work done by external agent = U3 - U2

work done by external agent = 3.11 *10^-9 - 1.038 *10^-9

work done by external agent = 2.072 *10^-9 J

the work done by external agent is 2.072 *10^-9 J

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