Question

(a) How much charge is on each plate of a 4.00 µF capacitor when it is connected to a 9.00 V battery?
1 µC
(b) If this same capacitor is connected to a 12.0 V battery, what charge is stored?
2 µC

Answer 1

a)

C = Capacitance = 4 uF = 4 x 10^-6 F

V = Voltage = 9 Volts

Q = charge stored = CV = (4 x 10^-6 ) (9) = 36 x 10^-6 = 36 uC

b)

C = Capacitance = 4 uF = 4 x 10^-6 F

V = Voltage = 12 Volts

Q = charge stored = CV = (4 x 10^-6 ) (12) = 48 x 10^-6 = 48 uC