**A Simple Network of Capacitors**

Part A = 80.0 μC

Part B = 80.0 μC

Part B = 37.3 V

**Solutions Below:**

_{1}= 6.00 μF, C

_{2}= 3.00 μF, C

_{3}= 5.00 μF. 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 μC.

Part A

_{1}on capacitor C

_{1}?

Start with the formula Q = CV. Also note that C_{1} and C_{2} are in parallel – so the voltage on each is the same:

Q = CV

Q_{1} = C_{1} * V

V = Q_{1} / C_{1}

and:

Q_{2} = C_{2} * V

V = Q_{2} / C_{2}

So:

Q_{1} / C_{1} = Q_{2} / C_{2}

Q_{1} / 0.000006 = 40 / 0.000003

0.5 * Q_{1} = 40

Q_{1} = 80

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

80.0 μC

Part B

_{3}?

Express your answer in microcoulombs to three significant figures.

The total charge at C_{3} will be the sum of the charges from C_{1} and C_{2}. Q_{1} was found in Part A, and Q_{2} was given:

Q_{3} = Q_{1} + Q_{2}

Q_{3} = 80 + 40

Q_{3} = 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

_{ab}?

Express your answer in microcoulombs to three significant figures.

Start by finding the equivalent capacitance C_{eq} for C_{1} and C_{2} 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 C_{eq} (from C_{1} and C_{2}) and C_{3} are in series, the total capacitance is given by:

C_{total} = (1 / C_{eq} + 1 / C_{3})^-1

C_{total} = (1 / 9 + 1 / 5)^-1

C_{total} = (0.311)^-1

C_{total} = 3.2143

Next using the formula Q = CV and figure out the voltage on C_{3}. Remember that since C_{eq} and C_{3} are in series, the charge will be the same on each of them. The charge (Q) for C_{eq} is the sum of the charges on C_{1} and C_{2}:

Q = CV

(80 + 40) = 3.2143 * V

120 = 3.2143V

V = 37.3

37.3 V