Two capacitors of values 6.00 mF and 9.00 mF are connected in series to a 30.0 V power supply. What is the resulting charge on the 6.00 mF capacitor?
A) 180 mC
B) 270 mC
C) 90.0 mC
D) 108 mC
E) 200 mC

Question

Quizplus · Accepted Answer

The equivalent capacitance of two capacitors in series is given by:

1/Ceq = 1/C1 + 1/C2

Substituting values, we get:

1/Ceq = 1/6.00×10⁻³ + 1/9.00×10⁻³
Ceq = 3.60×10⁻³ F

The charge on each capacitor will be the same since they are in series. The total charge supplied by the power supply will be:

Q = CV = Ceq × 30.0

Substituting values, we get:

Q = (3.60×10⁻³) × 30.0
Q = 0.108 C = 108 mC

Two Capacitors of Values 6