A 2.85 F capacitor is charged to 490 V and a 3.80 F capacitor is charged to 525

Question

A 2.85 F capacitor is charged to 490 V and a 3.80 F capacitor is charged to 525 V .

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.)

B) What will be the charge on each capacitor? (Enter your answers numerically separated by a comma.)

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.)

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

Given

Capacitance, C = 2.85 µF = 2.85 x 10 -6 F

Voltage, V = 490 V

Charge on first capacitor Q = CV = 1396.5 x 10 -6 coulomb

          C ' = 3.80 µF = 3.80 x 10 -6 F

          V ' = 525 V

Charge on second capacitor Q ' = C'V' = 1995 x 10 -6 coulomb

Since plates are in parallel, so

Resultant capacitance C " = C + C' = 6.65 x 10 -6 F

Total charge Q" = Q + Q' = 3391.5 x 10 -6 coulomb

The potential difference across each and the charge on each V" = Q"/C" = 510 volt

Charge on first capacitor q = CV" = 1453.5 x 10 -6 coulomb

Charge on second capacitor q ' = C'V" = 1938 x 10 -6 coulomb

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