# Physics: Potential Difference of a Capacitor

Please solve and explain problem.

Part 1:

A 0.410 mF (C1) and a 0.630 mF (C2) capacitor are connected in series to a 12.00 V battery. Calculate the potential difference across C1.

Part 2:

Calculate the charge on C2.

Part 3:

Now assume that the two capacitors are connected in parallel. Calculate the voltage difference across C1.

Part 4:

Calculate the charge on C2.

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## SOLUTION This solution is **FREE** courtesy of BrainMass!

1. The total capacitance in series is:

C = (0.41mF)(0.63mF)/(0.41mF+0.63mF)

C = 0.248mF

The total stored charge is:

q = CVbatt

q = (0.248mF)(12V)

q = 2.976mC

The voltage at C1 is:

V1 = q/C1

V1 = 2.976mC / 0.410mF

V1 = 7.259 V

2. Since the charge in series is constant, and we have solved it in the previous problem, the charge at C2 is:

q2 = q = 2.976mC

3. When the capacitors are connected in parallel, the voltage between 2 capacitors are constant. So we have:

V1 = Vbatt = 12V

4. The charge across C2 is:

q2 = C2 Vbatt

q2 = (0.630mF)(12V)

q2 = 7.56mC

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