Mastering Physics: Exercise 25.23

Mastering Physics: Exercise 25.23

A current-carrying gold wire has a diameter of 0.86 mm . The electric field in the wire is 0.51 V/m .

Part A

Question: What is the current carried by the wire?

Answer: 

p (resistivity) of Au = 2.4×10^(−8)

A = π × r^2

A = π × ((0.86×10^(−3) / 2)^2)

A = 5.8088 ×10^(−7)

p = E / J = E / (I / A) = EA / I

=> I = E×A / I

I = 0.51 × 5.8088 ×10^(−7) / (2.4×10^(−8))

I =  12  A

Mastering Physics: Resistance and Geometry

Mastering Physics: Resistance and Geometry

Part A

Question: Which conductor in the table has the greatest resistance?

Answer: E

Mastering Physics: Exercise 25.5

Mastering Physics: Exercise 25.5

Copper has 8.5×10^(28) free electrons per cubic meter. A 71.0 cm length of 12-gauge copper wire, that is 2.05 mm in diameter, carries 4.80 A of current.

Part A

Question: How much time does it take for an electron to travel the length of the wire?

Answer: 

I = n × |q| × v_d × A
v_d = I / (n × |q| × A)
v_d = 4.80 / (8.5×10^(28) 3.57×10^(−6) × 1.602×10^(−19) × π × ((3.57×10^(−3) / 2)^2) )
v_d = 1.06798 ×10^(−4)

t = L / v_d
t = 0.71 / (1.06798 ×10^(−4))

t =  6650  s  

Mastering Physics: PhET Tutorial: Circuit Construction Kit – Ohm’s Law and Power

Mastering Physics: PhET Tutorial: Circuit Construction Kit - Ohm's Law and Power

Part A

Question: Drag a battery into the construction panel, and use the voltmeter to determine which end of the battery is the positive terminal. The positive terminal has a higher potential than the negative terminal (recall that the voltmeter measures the potential difference between the red probe and the black probe).Which end of the battery is the positive terminal?

Answer:  the black end

Mastering Physics: Electricity and Water Analogy

Mastering Physics: Electricity and Water Analogy

Figure 1

Part A

Question: (Figure 1) Consider the following water circuit: water is continually pumped to high pressure by a pump, and then funnelled into a pipe that has lower pressure at its far end (else the water would not flow through the pipe) and back to the pump. Two such circuits are identical, except for one difference: the pipes in one circuit have a larger diameter than the pipes in the other circuit. Through which circuit is the flow of water greater?

Answer: Large pipe

Mastering Physics: Exercise 24.38

Mastering Physics: Exercise 24.38

A budding electronics hobbyist wants to make a simple 1.3 nF capacitor for tuning her crystal radio, using two sheets of aluminum foil as plates, with a few sheets of paper between them as a dielectric. The paper has a dielectric constant of 4.6, and the thickness of one sheet of it is 0.18 mm .

Part A

Question: If the sheets of paper measure 28 cm × 38 cm and she cuts the aluminum foil to the same dimensions, how many sheets of paper should she use between her plates to get the proper capacitance?

Answer: 

Area = 28 cm × 38 cm

Area = 1064 cm^2

Area = 0.1064 m^2

C = K× ϵ_0 × A / d

d = K× ϵ_0 × A / C

d = 4.6 × 8.85×10^(−12) × 0.1064 / (1.3×10^(−9))

d = 3.332×10^(−3)

Sheets = 3.332×10^(−3)  (0.18×10^(−3))

Sheets = 18.5

n =  19 sheets 

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  

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)

Mastering Physics: Exercise 24.3

Mastering Physics: Exercise 24.3

A parallel-plate air capacitor with a capacitance of 246 pF has a charge of magnitude 0.146 μC on each plate. The plates have a separation of 0.316 mm .

Part A

Question: What is the potential difference between the plates?

Answer: 

C = Q / V

=> V = Q / C

V = 0.146 ×10^(−6) / (246 ×10^(−12) )

V =  593  V  

Mastering Physics: The Capacitor as an Energy-Storing Device

Mastering Physics: The Capacitor as an Energy-Storing Device

An air-filled parallel-plate capacitor has plate area A and plate separation d. The capacitor is connected to a battery that creates a constant voltage V.

Part A

Question: Find the energy U_0 stored in the capacitor. Express your answer in terms of A, d, V, and ϵ_0. Remember to enter ϵ_0 as epsilon_0.

Answer: U_0 =  ϵ_0 × A × (V)^2 / 2d