(Figure 1) Part A Find the \"total charge\" stored in this network. Now, the eac

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(Figure 1)

Part A

Find the "total charge" stored in this network. Now, the each capacitor has a positive plate and a negative plate with equal but opposite charges. So if we added together all the charges of all the plates, we would always get zero. When we say "total charge" then, we mean adding together the Q 's for all the capacitors. This sum is equal to the total amount of positive charge we have on all our positive plates together, and also equal in magnitude to the total amount of negative charge stored on all the negative plates.

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Part B

Find the charge on each capacitor.

Enter your answer as two numbers, separated with a comma.

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Part C

Find the total energy stored in the network.

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Part D

Find the energy stored in each capacitor.

Enter your answers numerically separated by a comma.

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Part E

Find the potential difference across each capacitor.

Enter your answers numerically separated by a comma.

Consider the combination of capacitors shown below. The potential between points a and b is   28.7 V   .

(Figure 1)

Part A

Find the "total charge" stored in this network. Now, the each capacitor has a positive plate and a negative plate with equal but opposite charges. So if we added together all the charges of all the plates, we would always get zero. When we say "total charge" then, we mean adding together the Q 's for all the capacitors. This sum is equal to the total amount of positive charge we have on all our positive plates together, and also equal in magnitude to the total amount of negative charge stored on all the negative plates.

Q = ?C

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Part B

Find the charge on each capacitor.

Enter your answer as two numbers, separated with a comma.

Q35nF,Q75nF = ?C

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Part C

Find the total energy stored in the network.

U = ?J

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Part D

Find the energy stored in each capacitor.

Enter your answers numerically separated by a comma.

U35nF,U75nF = ?J

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Part E

Find the potential difference across each capacitor.

Enter your answers numerically separated by a comma.

V35nF,V75nF = V

Explanation / Answer

Part A)    total charge stored in this network = (35 + 75) * 10-9 * 28.7

                                                                       = 3.157   C

Part B)     charge on 35 nF capacitor = 35 * 10-9 * 28.7

                                                             = 1.0045 C

                charge on 75 nF capacitor = 75 * 10-9 * 28.7

                                                             = 2.1525 C

Part C)    total energy stored = 0.5 * 110 * 10-9 * 28.7 *28.7

                                                = 45.303   J

Part D)   Energy stored in 35 nF = 0.5 * 35 * 10-9 * 28.7 *28.7

                                                     = 14.414 J

                 Energy stored in 75 nF = 0.5 * 75 * 10-9 * 28.7 *28.7

                                                       = 30.888 J

Part E)    the potential difference across 35 nF = 28.7 V

                the potential difference across 75 nF = 28.7 V

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