Mastering Physics: Exercise 24.30

Mastering Physics: Exercise 24.30

For the capacitor network shown in the Figure (Figure 1) , the potential difference across ab is 36 V .

Figure 1

Part A

Question: Find the total charge stored in this network.

Answer: 

1 / C_ab = 1 / C1 + 1 / C2

1 / C_ab = 1 / (150 ×10^(−9) ) + 1 / (120 ×10^(−9))

1 / C_ab = 15000000

=> C_ab = 1 / 15000000

C_ab = 6.67 ×10^(−8)

Q = C×V

Q = 36 × 6.67 ×10^(−8)

Q = 2.4 ×10^(−6) C

Q =  2.4  μC  

Part B

Question: Find the charge on each capacitor.

Answer: 

Q_150nF = Q_120nF = 2.4  μC  

Q_150nF, Q_120nF =  2.4, 2.4  μC  

Part C

Question: Find the total energy stored in the network.

Answer: 

U = Q^2 = 2×C

U = (2.4×10^(−6))^2 / (2× 6.67×10^(−8))

U = 4.3×10^(−5) J

U =  43  μJ  

Part D

Question: Find the energy stored in each capacitor.

Answer: 

U_150nF = (2.4 ×10^(−6))^2 / (2× 150×10^(−9))

U_150nF = 1.9 ×10^(−5) J

U_150nF = 19 μJ

U_120nF = (2.4×10^(−6))^2 / (2× 120×10^(−9))

U_120nF = 2.4 ×10^(−5) J

U_120nF = 24 μJ

  

U_150nF, U_120nF =  19, 24  μJ

Part E

Question: Find the potential difference across each capacitor.

Answer: 
V_150nF = 2.4×10^(−6) / (150×10^(−9))
V_150nF = 16 V

V_120nF = 2.4×10^(−6) / (120×10^(−9))
V_120nF = 24 V

V_150nF, V_120nF =  16, 20  V