FIGURE 21-18
-A 4.0 μF and a 6.0 μF capacitor are connected in series across an 8.0-V DC source. What is the charge on the 6.0 μF capacitor?
A) 2.0 μC
B) 4.0 μC
C) 12 μC
D) 25 μC
E) 19 μC

Question

FIGURE 21-18 
-A 4.0 μF and a 6.0 μF capacitor are connected in series across an 8.0-V DC source. What is the charge on the 6.0 μF capacitor?
A) 2.0 μC 
B) 4.0 μC 
C) 12 μC 
D) 25 μC 
E) 19 μC

Quizplus · Accepted Answer

The total capacitance in the circuit can be found using the formula

1/C_total = 1/C1 + 1/C2

Substituting the values, we get

1/C_total = 1/4 + 1/6

1/C_total = 5/12

C_total = 12/5 μF

The total charge on the capacitors is given by

Q = CV

where C is the total capacitance and V is the voltage across the capacitors.

Substituting the values, we get

Q = (12/5) x 8

Q = 96/5 μC

Since the two capacitors are in series, they have the same charge. Thus,

Q1 = Q2 = 96/5 x (6/10) = 19.2/5

Q2 = 3.84 μC

Therefore, the answer is E) 19 μC (rounded to the nearest whole number).

FIGURE 21-18 -A 4.0 μF and a 6.0 μF Capacitor Are Connected