## Mastering Physics Solutions: A Simple Network of Capacitors

A Simple Network of Capacitors

Part A = 80.0 μC
Part B = 80.0 μC
Part B = 37.3 V

Solutions Below:

In the figure are shown three capacitors with capacitances C1 = 6.00 μF, C2 = 3.00 μF, C3 = 5.00 μF. The capacitor network is connected to an applied potential Vab. After the charges on the capacitors have reached their final values, the charge Q2 on the second capacitor is 40.0 μC.

Part A

What is the charge Q1 on capacitor C1?

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

Remember the units are in μC for this problem and to use 3 significant figures:

80.0 μC

Part B

What is the charge on capacitor C3?

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

Note that Mastering Physics considers this 3 significant figures. I consider Mastering Physics to not understand what a significant figure is.

120 μC

Part C

What is the applied voltage, Vab?

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 μC.

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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