A 3.00 F capacitor is charged to 480 V and a 3.55 F capacitor is charged to 500

Question

A 3.00 F capacitor is charged to 480 V and a 3.55 F capacitor is charged to 500 V .

Part A

These capacitors are then disconnected from their batteries, and the positive plates are now connected to each other and the negative plates are connected to each other. What will be the potential difference across each capacitor?

Enter your answers numerically separated by a comma.

Part B

What will be the charge on each capacitor?

Enter your answers numerically separated by a comma.

Part C

What is the voltage for each capacitor if plates of opposite sign are connected?

Express your answers using two significant figures. Enter your answers numerically separated by a comma.

Part D

What is the charge on each capacitor if plates of opposite sign are connected?

Express your answers using two significant figures. Enter your answers numerically separated by a comma.

Explanation / Answer

The first thing that will happen when you connect the two is a large spark as the voltages equalize.
a) correct, charge is conserved, and energy is NOT conserved.

Charge on the two caps is Q = CV
Q = 3x480 + 3.55x500 = 3215 µC

after they are connected, C = 6.55µF, so V = Q/C
V = 3215 µC / 6.55µF = 490.8 volts. (same voltage on both, obviously)

Q(3) = CV = 490.8x3 = 1472.4 uC
Q(3.55) = CV = 490.8x3.55 = 1742.34 uC

now you do it over but connect them opposing?
Now the charges subtract, Q = 3x480 - 3.55x500 = 335 uC

you can do the rest, it just repeats.

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