A 2.50- F capacitor is charged to 747 V and a6.80- F capacitor is charged to 580

Question

A 2.50-F capacitor is charged to 747 V and a6.80-F capacitor is charged to 580 V . These capacitors are then disconnected from their batteries. Next the positive plates are connected to each other and the negative plates are connected to each other. What will be the potential difference across each and the charge on each? [Hint: Charge is conserved.]

Part A

Determine the potential difference across the first capacitor.

Express your answer using three significant figures and include the appropriate units.

Part B

Determine the potential difference across the second capacitor.

Express your answer using three significant figures and include the appropriate units.

Part C

Determine the charge across the first capacitor.

Express your answer using three significant figures and include the appropriate units

Part D

Determine the charge across the second capacitor.

Express your answer using three significant figures and include the appropriate units.

due tomorrow!

Explanation / Answer

After making the given connections, the charge on the capacitors is conserved, that is

Charge before connections = charge after connections

C1V1 + C2V2 = (C1 +C2)V

so V = [C1V1 + C2V2]/(C1 +C2)

so substituing the values, C1 = 2.5E-6F; C2 = 6.8E-6F; V1 = 747V and V2 = 580V, we get

V = 624.89V

and Q1 = C1 V = 1.56mC

& Q2 =C2V = 4.25mC

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