30: Introduction to Electric Current

INTRO:
To understand the nature of electric current and the conditions under which it exists.
Electric current is defined as the motion of electric charge through a conductor. Conductors are materials that contain movable charged particles. In metals, the most commonly used conductors, such charged particles are electrons. The more electrons that pass through a cross section of a conductor per second, the greater the current. The conventional definition of current is

I=Q_total/Δt

where I is the current in a conductor and Qtotalis the total charge passing through a cross section of the conductor during the time interval Δt.

The motion of free electrons in metals not subjected to an electric field is random: Even though the electrons move fairly rapidly, the net result of such motion is that Q_total=0 (i.e., equal numbers of electrons pass through the cross section in opposite directions). However, when an electric field is imposed, the electrons continue in their random motion, but in addition, they tend to move in the direction of the force applied by the electric field.

In summary, the two conditions for electric current in a material are the presence of movable charged particles in the material and the presence of an electric field.

Quantitatively, the motion of electrons under the influence of an electric field is described by the drift speed, which tends to be much smaller than the speed of the random motion of the electrons. The number of electrons passing through a cross section of a conductor depends on the drift speed (which, in turn, is determined by both the microscopic structure of the material and the electric field) and the cross-sectional area of the conductor.

In this problem, you will be offered several conceptual questions that will help you gain an understanding of electric current in metals.
-----------------------------------------------------------------------------------------------------
PART A:
You are presented with several long cylinders made of different materials. Which of them are likely to be good conductors of electric current?

• copper
• aluminum
• glass
• quartz
• cork
• plywood
• table salt
• gold

SOLUTION:
As stated in the intro, metals are most likely to be good conductors of electric current,
so:
copper, aluminum, and gold
-----------------------------------------------------------------------------------------------------
PART B:
Metals are good conductors of electric current for which of the following reasons?

• They possess high concentrations of protons
• They possess low concentrations of protons
• They possess high concentrations of free electrons
• They possess low concentrations of free electrons

SOLUTION:
The intro states that "In metals, the most commonly used conductors, such charged particles are electrons."
so, the third option: They possess high concentrations of free electrons
-----------------------------------------------------------------------------------------------------
PART C:
Which of the following is the most likely drift speed of the electrons in the filament of a light bulb?

• 10^-8 m/s
• 10^-4 m/s
• 10 m/s
• 10⁴ m/s
• 10⁸ m/s

SOLUTION:
The second option,  10^-4 m/s

-----------------------------------------------------------------------------------------------------
PART D:
You are presented with several wires made of the same conducting material. The radius and drift speed are given for each wire in terms of some unknown units r and v. Rank the wires in order of decreasing electron current.
Rank from most to least electron current. To rank items as equivalent, overlap them.

SOLUTION:
Since the wires are made of the same material, the charge carriers and their densities are the same for all the wires.

Other conditions being equal, the current is proportional to the product of the cross-sectional area of the wire and the drift velocity, that is,

I=n|q|v_dA,

where I is the current, v_d is the drift velocity, A is the cross-sectional area, n is the density of charge carriers, and q is the charge on the carriers.

Therefore:

-----------------------------------------------------------------------------------------------------
PART E:
The drift speed of the electrons in a wire depends strongly on which of the following factors?

• The cross-sectional area of the wire
• The mass of the wire
• The temperature of the wire
• The internal electric field in the wire

SOLUTION:
In the intro, it states "the motion of electrons under the influence of an electric field is described by the drift speed"
so, the final option is correct: The internal electric field in the wire
-----------------------------------------------------------------------------------------------------
PART F:
What quality must the charge density on the surface of a conducting wire possess if an electric field is to act on the negatively charged electrons inside the wire?

The charge density must be... 

• positive
• negative
• nonuniform
• uniform

SOLUTION:
nonuniform

28: Problem 28.19

INTRO:
A 3.4-cm-diameter parallel-plate capacitor has a 1.6 mm spacing. The electric field strength inside the capacitor is 8.0×10⁴ V/m .
-----------------------------------------------------------------------------------------------------
PART A:
What is the potential difference across the capacitor?
Express your answer to two significant figures and include the appropriate units.

SOLUTION:

My favorite first step,​

Givens/ conversions:
diameter ≡ d = 3.4 cm = 0.034 m
spacing ≡ s = 1.6 mm = 0.0016 m
Electric field inside ≡ E = 8.0×10⁴ V/m

using equation 28.5, for the electric potential inside a parallel-plate capacitor: V = Es

so, V = (8.0×10⁴ V/m)(0.0016 m) = 128 V

the frustrating kicker here, it got me anyway, is that the instructions say to express your solution to two significant figures. 
128 has 3 sigfigs. annoying. You have to round up to 130...

⇒ ΔV_C = 130 V

-----------------------------------------------------------------------------------------------------
PART B:
How much charge is on each plate?
Express your answer to two significant figures and include the appropriate units.

SOLUTION:
An electric field is considered to be a surface charge density (with units of C/m²) divided by the vacuum permittivity constant (with units C/Vm). This means that the surface charge density, η, can be calculated by multiplying the electric field by the vacuum permittivity constant. 

η=E⋅ε₀​

the surface charge density is basically the charge distributed over an area, or η = Q/A
therefore, to solve for the charge, 

Q = ηA = E⋅ε₀⋅A
​

The area is just the area of the circular plate, or A = πr² = π/4 ⋅ d²

∴ Q = E⋅ε₀⋅π/4⋅d²

= (8.0×10⁴ V/m)(π/4)(8.854×10^-12 C/Vm)(0.034 m)²
⇒ Q = 6.431×10^-10 C​

but don't forget sigfigs... it only wants two...

⇒ Q = 6.4×10^-10 C​

27: Problem 27.50

INTRO:(Figure 1) shows two very large slabs of metal that are parallel and distance l apart. The top and bottom surface of each slab has surface area A. The thickness of each slab is so small in comparison to its lateral dimensions that the surface area around the sides is negligible. Metal 1 has total charge Q₁=Q and metal 2 has total charge Q₂=2Q. Assume Q is positive.

Figure 1

-----------------------------------------------------------------------------------------------------
PART A:
Determine the electric field strength E₁ in region 1. Give your answer as a multiple of Q/ε₀.
Express your answer in terms of some or all of the variables l, A, Q, and the constant π.

SOLUTION:
One Gaussian surface is a cylinder with end-area a extending past the two metal slabs. A second Gaussian surface ends in region 3, the space between the slabs

Using Gauss's Law, ∫E⋅dA = ∫_top E⋅dA + ∫_bottom E⋅dA + ∫_sides E⋅dA = E₁a + E₅a + 0 = Q_in / ε₀

The total charge per unit area on both surfaces of the top slab is Q₁/A = Q/A, so the charge enclosed within the cylinder is Qa/A.

Similarly, the enclosed charge on the lower slab is Q₂a/A = 2Qa/A

Thus, E₁a + E₅a = Qa/(A⋅ε₀) + 2Qa/(A⋅ε₀) = 3Qa/(A⋅ε₀)
∴ E₁ + E₅ = 3Q/(A⋅ε₀)

Fields E₁ & E₅ are both a superposition of the fields of four sheets of surface charge. Because the field of a plane of charge is independent of distance from the plane, the superposition at points above the top plane must be the same magnitude, but opposite direction, as the superposition at points below the bottom plane. 

∴ E₁ = E₅ = 1/2(3Q/(A⋅ε₀))

E₁ = 3/2A ⋅Q/ε₀

-----------------------------------------------------------------------------------------------------
PART B:
Determine the electric field strength E₂ in region 2. Give your answer as a multiple of Q/ε_0.
Express your answer in terms of some or all of the variables l, A, Q, and the constant π.

SOLUTION:
Because these are metals we immediately know that E₂ = E₄ = 0
0

-----------------------------------------------------------------------------------------------------

PART C:

Determine the electric field strength E₃ in region 3. Give your answer as a multiple of Q/ε₀.
Express your answer in terms of some or all of the variables l, A, Q, and the constant π.

SOLUTION:

Consider the Gaussian surface on the right. The lower slab is more positive than the upper slab, so the electric field in region 3 must point upward, into the lower face of this cylinder.

∫E⋅dA = ∫_top E⋅dA + ∫_bottom E⋅dA + ∫_sides E⋅dA = E₁a - E₃a + 0 = Q_in / ε₀

where the minus sign with E₃ is because of the direction.  We know E₁, and we've already determined Q_in = Qa/A.

Thus, E₃ = E₁ - Q/(A⋅ε₀) = 3Q/(2A⋅ε₀) - Q/(A⋅ε₀) 

∴ E₃ = Q/(2A⋅ε₀)

E₃ = 1/2A ⋅Q/ε₀

-----------------------------------------------------------------------------------------------------

PART D:

Determine the electric field strength E₄ in region 4. Give your answer as a multiple of Q/ε₀.

Express your answer in terms of some or all of the variables l, A, Q, and the constant π.

SOLUTION:

See part B

0

-----------------------------------------------------------------------------------------------------

PART E:

Determine the electric field strength E₅ in region 5. Give your answer as a multiple of Q/ε₀.

Express your answer in terms of some or all of the variables l, A, Q, and the constant π.

SOLUTION:

See Part A

E₅ = 3/2A ⋅Q/ε₀

-----------------------------------------------------------------------------------------------------

PART F:

Determine the surface charge density η_a on the surface a. Give your answer as a multiple of Q/A.

SOLUTION:

The electric field at the surface of a conductor is E = η/ε₀

We can use the known fields and η = ε₀E to find the four surface charge densities.  

At surface a, E₁ points away from the surface, Thus

η_a =  ε₀E₁ = ε₀⋅3Q/(2A⋅ε₀) = 3Q/(2A)

η_a = 3/2⋅Q/A

-----------------------------------------------------------------------------------------------------

PART G:

Determine the surface charge density η_b on the surface b. Give your answer as a multiple of Q/A.

SOLUTION:

At surface b, E₃ points toward the surface, Thus

η_b =  -ε₀E₁ = -ε₀⋅Q/(2A⋅ε₀) = -Q/(2A)

η_b = -1/2⋅Q/A

-----------------------------------------------------------------------------------------------------

PART H:

Determine the surface charge density η_c on the surface c. Give your answer as a multiple of Q/A.

SOLUTION:

Surface c is opposite to surface b, because the field points away from the surface.
Therefore, η_c = - η_b = Q/(2A)

η_c = 1/2⋅Q/A

-----------------------------------------------------------------------------------------------------

PART I:

Determine the surface charge density η_d on the surface d. Give your answer as a multiple of Q/A.

SOLUTION:

At surface d, the field points away from the surface and has the same strength as E₁
Therefore, η_d = η_a = 3Q/(2A)

η_d = 3/2⋅Q/A

-----------------------------------------------------------------------------------------------------