In the figure are shown three capacitors with capacitances C_1 = 6.00 mu F, C_2

Question

In the figure are shown three capacitors with capacitances C_1 = 6.00 mu F, C_2 = 3.00mu F, C_3 = 5.00 mu F. The capacitor network is connected to an applied potential V_ab. After the charges on the capacitors have reached their final values, the charge Q_2 on the second capacitor is 40.0 mu C. What is the charge Q_1 on capacitor C_1? What is the charge on capacitor C$ ? Express your answer in microcoulombs to three significant figures. What is the applied voltage, V_ab ? Express your answer in volts to three significant figures.

Explanation / Answer

Start with the formula Q = CV. Also note that C1 and C2 are in parallel – so the voltage on each is the same:

Q = CV
Q1 = C1 * V
V = Q1 / C1

and:

Q2 = C2 * V
V = Q2 / C2

So:

Q1 / C1 = Q2 / C2
Q1 / 0.000006 = 40 / 0.000003
0.5 * Q1 = 40
Q1 = 80 ?C

B)

The total charge at C3 will be the sum of the charges from C1 and C2. Q1 was found in Part A, and Q2 was given:

Q3 = Q1 + Q2
Q3 = 80 + 40
Q3 = 120 ?C

C)

Start by finding the equivalent capacitance Ceq for C1 and C2 as a single unit. Since they are in parallel, you can just add the two capacitances together, and get 9.0 ?C.

Next, find the total capacitance of the entire circuit. Since Ceq (from C1 and C2) and C3 are in series, the total capacitance is given by:

Ctotal = (1 / Ceq + 1 / C3)^-1
Ctotal = (1 / 9 + 1 / 5)^-1
Ctotal = (0.311)^-1
Ctotal = 3.2143

Next using the formula Q = CV and figure out the voltage on C3. Remember that since Ceq and C3 are in series, the charge will be the same on each of them. The charge (Q) for Ceq is the sum of the charges on C1 and C2:

Q = CV
(80 + 40) = 3.2143 * V
120 = 3.2143V

V = 37.3

37.3 V

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