Verified answer

U = (1/2) C V^2. So before they are connected the second capacitor has zero stored energy.

U1 = (1/2) (2.07x10^-6F) (23V)^2

U1 = 5.48x10^-4 J

For (b), charges move until the voltage across each is identical, and the sum of the charges remains the same.

C = Q/V

Q = CV = (2.07x10^-6F) (23.0V)

Q = 4.76x10^-5 C

Let the fraction of charge on capacitor1 be x, so the fraction on capacitor 2 is (1-x).

V = Q/C

Q1/C1 = Q2/C2

x / C1 = (1-x) / C2

C2 x = C1 - C1x

x (C2 + C1) = C1

x = C1 / (C2 + C1)

x = 2.07 / (2.07 + 4.38)

x = 0.321

So after connecting Q1 = 0.321 * 4.76x10^-5C = 1.53 x 10^-5C

Q2 = (1 - 0.321) * 4.76x10^-5C = 3.23 x 10^-5C

They should each have the same voltage.

V1 = Q1 / C1 = 1.53x10^-5C / 2.07x10^-6F => 7.38V

V2 = Q2 / C2 = 3.23x10^-5C / 4.38x10^-6F => 7.38V

U = U1 + U2

U1 = .5 C1 (V)^2 = .5 2.07x10^-5F (7.38V)^2 => 5.63x10^-5 J

U2 = .5 C2 (V)^2 = .5 4.38x10^-5F (7.38V)^2 => 1.19x10^-4 J

U = 1.76x10^-4 J

(c)

dU = U1 - U2

dU = 5.48x10^-4J - 1.76x10^-4J

dU = 3.72x10^-4J

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