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

Mastering Physics: Exercise 23.29

Mastering Physics: Exercise 23.29

A uniformly charged thin ring has radius 14.5 cm and total charge 22.5 nC . An electron is placed on the ring's axis a distance 33.0 cm from the center of the ring and is constrained to stay on the axis of the ring. The electron is then released from rest.

Part A

Question: Describe the subsequent motion of the electron.

Answer: When the electron is on either side of the center of the ring, the ring exerts an attractive force directed toward the center of the ring. This restoring force produces oscillatory motion of the electron along the axis of the ring, with amplitude, which equals initial distance between the electron and the ring's center. The force on the electron is not of the form F=-k*x so the oscillatory motion is not simple harmonic motion.

Mastering Physics: Exercise 23.23

Mastering Physics: Exercise 23.23

Part A

Question: An electron is to be accelerated from a velocity of 3.50×10^6 m/s to a velocity of 8.50×10^6 m/s . Through what potential difference must the electron pass to accomplish this?

Answer: 

U1 − U2 = K2 − K1

K2 − K1 = 1/2 × 9.12×10^(−31) kg × ( (8.50 × 10^(6))^2 −  (3.50×10^(−6))^2 )

K2 − K1 = 2.733×10^(−17)

V = U / q

V1 − V2 = 2.733×10^(−17) / (-1.6 ×10^(−19))

V1 − V2 =  -171  V

Mastering Physics: Exercise 23.5

Mastering Physics: Exercise 23.5

A small metal sphere, carrying a net charge of q1 = -2.70 μC , is held in a stationary position by insulating supports. A second small metal sphere, with a net charge of q2 = -7.50 μC and mass 1.70 g , is projected toward q1. When the two spheres are 0.800 m apart, q2 is moving toward q1 with speed 22.0 m/s (Figure 1) . Assume that the two spheres can be treated as point charges. You can ignore the force of gravity.

Figure 1

Mastering Physics: Video Tutor: Charged Conductor with Teardrop Shape

Mastering Physics: Video Tutor: Charged Conductor with Teardrop Shape

Part A

Question: Two conducting spheres are each given a charge Q. The radius of the larger sphere is three times greater than that of the smaller sphere. If the electric field just outside of the smaller sphere is E_0, then the electric field just outside of the larger sphere is

Answer: 1/9 E_0

Mastering Physics: PhET Tutorial: Charges and Electric Potential

Mastering Physics: PhET Tutorial: Charges and Electric Potential

Part A

Question: Using the voltage meter, you should find that 1 m away from the charge, the voltage is 9 V.  What is the voltage 2 m away from the charge?

Answer: 4.5 V 

Part B

Question: What is the voltage 3 m away from the charge?

Answer:  3 V

Mastering Physics: Electric Fields and Equipotential Surfaces

Mastering Physics: Electric Fields and Equipotential Surfaces

The dashed lines in the diagram represent cross sections of equipotential surfaces drawn in 1 V increments. (Figure 1)

Figure 1

Part A

Question: What is the work done by the electric force to move a 1 C charge from A to B?

Answer: 0 J 

Mastering Physics: Relationship between Electric Force and Electric Potential Conceptual Question

Mastering Physics: Relationship between Electric Force and Electric Potential Conceptual Question

Three points (A, B, and C) are located on equipotential lines as shown. (Figure 1)

Figure 1

Part A

Question: A proton is released from Point A. Indicate the direction of the electric force vector acting on the proton.

Answer: The electric force vector at Point A  points to the left.

Mastering Physics: Electric Potential, Potential Energy, and Force

Mastering Physics: Electric Potential, Potential Energy, and Force

Part A

Question: Electric field lines always begin at _______ charges (or at infinity) and end at _______ charges (or at infinity). One could also say that the lines we use to represent an electric field indicate the direction in which a _______ test charge would initially move when released from rest. Which of the following fills in the three missing words correctly?

Answer:  (positive; negative; positive)

Mastering Physics: Electric Potential Energy versus Electric Potential

Mastering Physics: Electric Potential Energy versus Electric Potential

Part A

Question: Find the force F⃗ (z) on an object of mass m in the uniform gravitational field when it is at height z=0.

Answer: F⃗ (z) =  −mgk^

Part B

Question: Now find the gravitational potential energy U(z) of the object when it is at an arbitrary height z. Take zero potential to be at position z=0. Keep in mind that the potential energy is a scalar, not a vector.

Answer: U(z) =  mgz

Mastering Physics: Exercise 22.10

Mastering Physics: Exercise 22.10

A point charge q1 = 3.80 nC is located on the x-axis at x = 2.10 m , and a second point charge q2 = -6.20 nC is on the y-axis at y = 1.15 m .

Part A

Question: What is the total electric flux due to these two point charges through a spherical surface centered at the origin and with radius r1 = 0.555 m ?

Answer:  Φ =  0  N⋅m^2/C 

Mastering Physics: Exercise 22.3

Mastering Physics: Exercise 22.3

You measure an electric field of 1.32×10^6 N/C at a distance of 0.156 m from a point charge. There is no other source of electric field in the region other than this point charge.

Part A

Question: What is the electric flux through the surface of a sphere that has this charge at its center and that has radius 0.156 m ?

Answer: 

Φ =  E⋅A 

= 1.32×10^(6) N/C × 4π(r^2)

=1.32×10^(6) × 4π(0.156)^2

= 4.04×105  N⋅m^2/C  

Mastering Physics: Exercise 22.21

Mastering Physics: Exercise 22.21

A hollow, conducting sphere with an outer radius of 0.251 m and an inner radius of 0.201 m has a uniform surface charge density of +6.55×10^(−6) C/m^2 . A charge of -0.640 μC is now introduced into the cavity inside the sphere.

Part A

Question: What is the new charge density on the outside of the sphere?

Answer: 
σ = Q / A
6.55 × 10^(−6) = Q / (0.251)^2 × 4π

=> Q = 5.1856 ×10^(−6)

New Q = 5.1856 ×10^(−6)  − 0.64 ×10^(−6)
= 4.5456 ×10^(−6)

=> σ = Q / A
σ = 4.5456 ×10^(−6) / (0.251)^2 × 4π
σ =  5.74×10^(−6)  C/m^2  

Mastering Physics: Video Tutor: Electroscope in Conducting Shell

Mastering Physics: Video Tutor: Electroscope in Conducting Shell

Part A

Question: As in the video, we apply a charge +Q to the half-shell that carries the electroscope. This time, we also apply a charge –Q to the other half-shell. When we bring the two halves together, we observe that the electroscope discharges, just as in the video. What does the electroscope needle do when you separate the two half-shells again?

Answer: It does not deflect at all.

Mastering Physics: A Conducting Shell around a Conducting Rod

Mastering Physics: A Conducting Shell around a Conducting Rod

(Figure 1) An infinitely long conducting cylindrical rod with a positive charge λ per unit length is surrounded by a conducting cylindrical shell (which is also infinitely long) with a charge per unit length of −2λ and radius r_1, as shown in the figure.

Figure 1

Part A

Question: What is E(r), the radial component of the electric field between the rod and cylindrical shell as a function of the distance r from the axis of the cylindrical rod?

Answer: E(r) =  λ/(2πrϵ_0)

Mastering Physics: The Charge Inside a Conductor

Mastering Physics: The Charge Inside a Conductor

(Figure 1) A spherical cavity is hollowed out of the interior of a neutral conducting sphere. At the center of the cavity is a point charge, of positive charge q.

Figure 1

Part A

Question: What is the total surface charge q_int on the interior surface of the conductor (i.e., on the wall of the cavity)?

Answer: q_int =  −q

Mastering Physics: The Electric Field and Surface Charge at a Conductor

Mastering Physics: The Electric Field and Surface Charge at a Conductor?

A conductor is placed in an external electrostatic field. The external field is uniform before the conductor is placed within it. The conductor is completely isolated from any source of current or charge.

Part A

Question: Which of the following describes the electric field inside this conductor?

Answer: It is always zero.

Mastering Physics: Gauss's Law

Mastering Physics: Gauss's Law

To understand the meaning of the variables in Gauss's law, and the conditions under which the law is applicable.Gauss's law is usually written ΦE=∮E⃗ ⋅dA⃗ =q_encl/ϵ_0,where ϵ_0=8.85×10^(-12) C^2/(N⋅m^2) is the permittivity of vacuum.

Part A

Question: How should the integral in Gauss's law be evaluated?

Answer: over a closed surface

Mastering Physics: Video Tutor: Charged Rod and Aluminum Can

Mastering Physics: Video Tutor: Charged Rod and Aluminum Can

Part A

Question: Consider the situation in the figure below, where two charged rods are placed a distance d on either side of an aluminum can. What does the can do?

Answer: stays still