A 2.50- F capacitor is charged to 760 V and a 6.80- F capacitor is charged to 56

Question

A 2.50-F capacitor is charged to 760 V and a 6.80-F capacitor is charged to 567 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. V1=?

Part B

Determine the potential difference across the second capacitor. V2=?

Part C

Determine the charge across the first capacitor. Q1=?

Part D

Determine the charge across the second capacitor. Q2=?

Explanation / Answer

Initial charge on 2.50-F capacitor   = 2.50 * 760

                                                           = 1900 C

Initial charge on 6.80-F capacitor   = 6.80 * 567

                                                           = 3855.6 C

Part A)    the potential difference across the first capacitor. V1

           1900 + 3855.6 = (2.5 +6.8)V1

       => V1   = 618.88 V

Part B)    the potential difference across the second capacitor. V2

           1900 + 3855.6 = (2.5 +6.8)V2

       => V2 = 618.88 V

Part C)   charge across the first capacitor ,Q1 = 2.5 * 618.88

                                                                         = 1547.2 C

Part D)   charge across the second capacitor ,Q2 = 6.8 * 618.88

                                                                         = 4208.38 C

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