Two capacitors with capacitances of 1.0 and 0.50 mF,respectively,are connected in series.The system is connected to a 100-V battery.What charge accumulates on the 1.0-mF capacitor?
A)150 mC
B)100 mC
C)50 mC
D)33 mC

Question

Quizplus · Accepted Answer

Applying the formula for capacitors in series:
1/Ceq = 1/C1 + 1/C2
1/Ceq = 1/1.0x10^-3 + 1/0.50x10^-3
Ceq = 0.33x10^-3 F

Using the formula Q = CV:
Q = (0.50x10^-3)(100) = 50x10^-3 C
Therefore, the charge accumulated on the 1.0-mF capacitor is half of the charge on the 0.50-mF capacitor:
Q1 = 50x10^-3 / 2 = 25x10^-3 C or 25 mC. 
So, the answer is D) 33 mC, which is the closest value to 25 mC.

Two Capacitors with Capacitances of 1