A 3.00 μF capacitor is charged to 480 V and a 4.00 μF 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?

Q before connected = Q after connected C1V1+C2V2 = (C1+C2) V

C1 = 3*10^-6 F

V1 = 480v

C2 = 4*10^-6 F

V2 = 500v

(3*10^-6) * (480) + (4*10^-6) * (500) = (3*10^-6 + 4*10^-6) * V

Simplifying the above, we get:

(1440 * 10^-6) + (2000 * 10^-6) = (7 * 10^-6) * V.

Further simplified as:

3440 * 10^-6 = 7 * 10^-6 * V

Making V the subject

V = 491.43volts

Therefore the potential difference across each capacitor is 491.43v