Mastering Physics: Exercise 24.17

Mastering Physics: Exercise 24.17

In the figure (Figure 1) , each capacitor has 4.60 μF and V_ab = 31.0 V .

Figure 1

Question: Calculate the charge on each capacitor.

Answer: 

1 / C_12 = (1 / C_1) + (1 / C_2)

1 / C_12 = (1 / (4.60×10^(−6) ) + (1 / (4.60×10^(−6) )

1 / C_12 = 434782.6087

=> C_12 = 1 / (434782.6087)

C_12 = 2.3 ×10^(−6)

C_123  = C_12 + C_3

C_123 = 2.3 ×10^(−6) + 4.6 ×10^(−6)

C_123 = 6.9 ×10^(−6)

1 / C_1234 = C_123 + C_4

1 / C_1234 = (1/ (6.9 ×10^(−6)) + (1 / (4.6 ×10^(−6))

1 / C_1234 = 362318.84

=> C_1234 = 1 / (362318.84)

C_1234 = 2.76 ×10^(−6)

Q = C×V

Q = 2.76 ×10^(−6) × 31

Q = 8.556 ×10^(−5)

Q = Q_4 = Q_123 = 8.556 ×10^(−5)

Part D answer: 8.556 ×10^(−5)

V4 = Q_4 / C

V4 = 8.556 ×10^(−5) / 4.6 ×10^(−6)

V4 = 18.6 V

Part H answer: 18.6 V

V_123 = V_123 / C

V_123 = 8.556 ×10^(−5) / (6.9 ×10^(−6))

V_123 = 12.4 V

V_ad = V_123 = 12.4 V

Part I answer: 12.4 V

V_12  = V_3 = 12.4 V

Part G answer: 12.4 V

V_12 = V_1 + V_2

V_1 = V_2 = V_12 / 2

V_1 = V_2 = 12.4 / 2

V_1 = V_2 = 6.2 V

Part E answer: 6.2 V

Part F answer: 6.2 V

Q_3 = V_3 × C

Q_3 = 12.4 × 4.6 ×10^(−6)

Q_3 = 5.70 ×10^(−5)

Part C answer: 5.70 ×10^(−5)

Q_12 = V_12 × C

Q_12 = (12.4) × 2.3 ×10^(−6)

Q_12 = 2.85 ×10^(−5)

Q_12 = Q_1 = Q_2 = 2.85 ×10^(−5)

Part A answer: 2.85 ×10^(−5)

Part B answer: 2.85 ×10^(−5)

Part A

Answer: 

Q1 =  2.85×10^(−5)  C  

Part B

Answer: Q2 =  2.85×10^(−5)  C

Part C

Answer: Q3 =  5.70×10^(−5)  C 

Part D

Answer: Q4 =  8.56×10−5  C  

Question: Calculate the potential difference across each capacitor.

Part E

Answer: V1 =  6.20  V  

Part F

Answer: V2 =  6.20  V  

Part G

Answer: V3 =  12.4  V

Part H

Answer: V4 =  18.6  V  

Part I

Question: Calculate the potential difference between points a and d.

Answer: V_ad =  12.4  V