26: PSS 26.1: The Electric Field of Multiple Point Charges

INTRO:
Three positively charged particles, with charges q₁=q, q₂=2q, and q₃=q (where q>0), are located at the corners of a square with sides of length d. The charge q₂ is located diagonally from the remaining (empty) corner.

Find the magnitude of the resultant electric field Enet in the empty corner of the square.

-----------------------------------------------------------------------------------------------------
PART A:
Below is an incomplete pictorial representation of the situation described in this problem. Complete the sketch by drawing the electric field due to each charge at point P. Make sure that all your vectors have the correct orientation.
The orientation of your vectors will be graded. The length of your vectors will not be graded.

SOLUTION:
Very obvious. The direction of the electric field is the same as the connecting line between the charge and point P

-----------------------------------------------------------------------------------------------------
PART B:
Now, consider the resultant electric field E_net at P. With reference to the coordinate system shown in the previous part, which component of E_net, if any, is zero in this problem?

• only the x component
• only the y component
• both the x and y components
• neither the x nor the y component

SOLUTION:
none of the components is zero, so the last option is correct
neither the x nor the y component

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

PART C:

Determine the magnitude E_net of the net electric field at point P. Use K for the electrostatic constant.
Express your answer in terms of q, d, and K.

SOLUTION:

Using equation (26.1) E = kq/r², all components of all electric fields can be determined. 

the electric field due to q₃ at point P is just an x-component of a field...

the distance in consideration is just d, so
E₃ = E_3x =kq/d²

the electric field due to q₂ at point P is both an x-component and a y-component of a field...

since the setup is that of a square, the magnitude of the x-component is equal to the magnitude of the y-component. 
E_2x = E₂cos(45) = E₂cos(45) = E₂⋅1/√2 

E_2x = kq√2/(2d²)

the y-component is the same but in the y-direction.

E_netx = E_3x + E_2x 

= kq/d² + (√2/2)kq/d²
⇒E_netx =(1+√2/2)kq/d²

the y-magnitude of this net electric field is the same as the x-magnitude, but in the y-direction. 
total mag = √(x+y)

& if |x| = |y| ... then total mag = x⋅√2 = y⋅√2

→ E_net = √2⋅(1+√2/2)kq/d²

⇒ E_net =(1+√2)kq/d²

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

PART D:

Intuitively, which of the following would happen to E_net if d became very large?

1. A) E_net should reduce to the field of a point charge of magnitude q.
2. B) E_net should reduce to the field of a point charge of magnitude 4q.
3. C) The larger d becomes, the smaller the magnitude of E_net will be.
4. D) The larger d becomes, the greater the magnitude of E_net will be.

Enter the letters of all the correct answers in alphabetical order. Do not use commas. For instance, if you think that A and D are correct, enter AD.

SOLUTION:

C) The larger d becomes, the smaller the magnitude of E_net will be

NOTE:

Looking at your answer derived in Part C, you should see that the magnitude Enet will decrease as d gets larger, just as you would expect. Your results do make sense! It is interesting to notice that the field will not reduce to that of a point charge (of magnitude 4q) unless the point P is moved farther away from the three charges while they themselves remain at their original positions.

Chapter 30: Notes Sheet

30.1 The Electron Current

Some vocabulary/ concepts:

- Current is the flow of charge through a conductor. 
- Charge carriers are the charges that move in a conductor.

Electrons are the charge carriers in metals​

- The electron drift speed, v_d, is the net motion caused by pushing electrons through a conductor with an electric field.
- The electron current, i_e, is the number of electrons per second that pass through a cross section of a conductor.
- The number of electrons, N_e, is literally the quantity of electrons that pass through the cross section.
- The number density, n_e, is the quantitative density of conduction electrons per volume in the conductor.

Equations:

N_e = i_e⋅Δt                                                 (30.1)
N_e = n_e⋅V = n_e⋅A⋅Δx = n_e⋅A⋅v_d⋅Δt          (30.2)
i_e = n_e⋅A⋅v_d                                             (30.3)

Tables/ Useful info:

30.2 Creating a Current

Some vocabulary/ concepts:

- An electron current is a nonequilibrium motion of charges sustained by an internal electric field.
- A nonuniform distribution of surface charges along a wire creates a net electric field inside the wire that points from the more positive end of the wire toward the more negative end of the wire. This is the internal electric field E that pushes the electron current through the wire. 
- Collisions transfer much of the electron's kinetic energy to the ion and thus to the thermal energy of the metal. This transfer is the 'friction' that raises the wire's temp.
- The magnitude of the electron's average velocity, due to the electric field, is the drift speed of the electron.
- The mean time between collisions, Δt, is designated as τ.
- The electron current is directly proportional to the electric field strength.

Equations:

v_d = e⋅τ⋅E/m                (30.7)
i_e = n_e⋅e⋅τ⋅A⋅E/m        (30.8)

30.3 Current and Current Density

Some vocabulary:

- The Ampere is the SI unit for current. It is equivalent to a Coulomb per second.
- The direction of current is defined to be the direction in which positive charges seem to move. - But the direction of the current I in a metal is opposite the direction of motion of the electrons
- The current density is current per square meter of cross section in the conductor
- The rate of electrons leaving a lightbulb (or any other device) is exactly the same as the rate of electrons entering the bulb. AKA, current doesn't change. → Law of conservation of current

Equations:

I = dQ/dt                                                   (30.9)
Q=I⋅Δt                                                      (30.10)

I = Q/Δt = eN_e/Δt = e i_e                             (30.11)
Current Density ≡ J = I/A = n_e⋅e⋅v_d          (30.13)
ΣI_in = ΣI_out                                                (30.14)

30.4 Conductivity and Resistivity

Some vocabulary/ concepts:

- The conductivity of a material is a characteristic independent of temperature or size. It is designated with the greek letter sigma, σ. it has units of Ω^-1m^-1
- The resistivity is the inverse of conductivity and is designated with the Greek letter rho, ρ. It has units of Ωm

Equations:

J = n_eev_d = n_ee²τE/m          (30.15)
σ = n_ee²τ/m                        (30.16)
J = σE                                  (30.17)
ρ = 1/σ = m/[n_ee²τ]            (30.18)

Tables/ Useful Info:

30.5 Resistance and Ohm's Law

Some vocabulary/ concepts:

- The current is proportional to the potential difference between the ends of a conductor.
- Resistance is a property of a conductor and depends on length, diameter, and resistivity of the material make-up. The units for resistance are ohms, Ω. 1 Ω = 1 Volt/Ampere = 1 V/A
- A battery is a source of potential difference.
- An ideal wire has no resistance, or R = 0 Ω
- Resistors are poor conductors, and are used to limit current within circuits. 
- Insulators are materials that barely conduct any electricity.
- An ideal insulator has infinite resistance, or R = ∞ Ω, and no electricity is conducted through it. 

Equations:

E = ΔV/Δs = ΔV/L         (30.19)
I = JA = AσE = AE/ρ      (30.20)
I = AΔV/(ρL)                 (30.21)
I = ΔV/R                       (30.23)

Ohm's Law: V = IR

30: Stretching the A-Rod

INTRO:
To learn to apply the concept of current density and microscopic Ohm's law.
A slab of metal of volume V is made into a rod of length L. The rod carries current I when the electric field inside is E.
-----------------------------------------------------------------------------------------------------
PART A:
Find the resistivity of the metal ρ.
Expess your answer in terms of the given quantities.

SOLUTION:
So, first off, V is volume, not voltage. So be careful to read the instructions. V = l⋅w⋅h = A⋅L & A = V/L

eq. (30.13) & eq. (30.17):
J = I/A = σE = E/ρ
→ ρ = E⋅A/I
sub A = V/L since those are given quantities... 
⇒ ρ = EV/IL

NOTE: I think it is hilarious that this solution spells EVIL. 
It's the little things... 
-----------------------------------------------------------------------------------------------------
PART B:
The rod is now stretched so that its length is doubled. If the electric field remains the same, what is the new current I′ in the rod?
Express your answer in terms of some or all of the quantities given in the problem introduction.

SOLUTION:

So, L₂ = 2L₁ 
if the electric field remains the same, then that means that the volume remains constant, or V₁ = V₂ (V = A⋅L)
∴ A₂⋅L₂ = A₁⋅L₁

→ A₂⋅2L₁ = A₁⋅L₁
→ A₂ = ½A₁​

but J₂ = J₁ 

so... I₂/A₂ = I₁/A₁

I₂ = I₁½A₁/A₁

⇒ I₂ = I₁/2, or

I' = I/2       (note this is capital i, as in current, and not one, 1)

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

PART C:

A piece of copper is made into a rod with a square cross-section. The side of the square is 2.00 centimeters. The resistivity of copper is 1.7⋅10⁻⁸ Ω⋅m. An unknown electric field E, directed along the rod, creates a current of 12.0 amperes through the rod. Find the magnitude of E.
Use two significant figures in your answer. Express your answer in newtons per coulomb.

SOLUTION:

Givens/ Conversions:​

b = h = 2 cm = 0.02 m ∴ A = 4×10^-4 m²
ρ = 1.7×10^-8 Ωm
I = 12 A

using eq. (30.20) ... I = AE/ρ ... 

E = Iρ/A

=(12 A)(1.7×10^-8 Ωm)/(4×10^-4 m²)
⇒ E = 5.10×10^-4 N/C
Wolfram LINK​

30: Current in a Wire

INTRO:
A metallic wire has a diameter of 4.12 mm. When the current in the wire is 8.00 A, the drift velocity is 5.40×10⁻⁵ m/s.
-----------------------------------------------------------------------------------------------------
PART A:
What is the density of free electrons in the metal?
Express your answer numerically in m⁻³ to two significant figures.

SOLUTION:

Givens/ Conversions:​

wire diameter ≡ d = 4.12 mm = 4.12×10^-3 m
Current ≡ I = 8 A
drift velocity ≡ v_d = 5.40×10⁻⁵ m/s.

Using eq. (30.13) ... J = I/A = n⋅e⋅v_d

→ n = I/(A⋅e⋅v_d)​

e = 1.6×10^-19 C
& A = π/4⋅d²

∴ n = I/(π/4⋅d²⋅e⋅v_d)

= (8 A)/(π/4⋅(4.12×10^-3 m)²⋅(1.6×10^-19 C)⋅(5.40×10⁻⁵ m/s))
⇒ n = 6.9×10²⁸ m^-3
Wolfram LINK

30: A Microscopic View of Resistivity

INTRO:
Recall that the density J of current flowing through a material can be written in terms of microscopic properties of the material: j=nqv_d, where n is the density of current carriers, q is the charge of one current carrier, and v_d is the drift velocity of a current carrier. In a metal, the current carriers are electrons.
-----------------------------------------------------------------------------------------------------
PART A:
The drift velocity is the component of the current-carrier's velocity due to acceleration from the electric field in the conductor. This corresponds to the average speed of all of the current carriers in the conductor. The current carriers also have random thermal motions, but the randomness causes the velocities due to thermal motion to cancel when averaged over a large number of current carriers. If the electric field inside of the conductor has magnitude E, and the charge q is accelerated from rest for a time τ, what is the final speed v of the charge?
Express the speed in terms of E, q, τ, and the mass m of the charge.

SOLUTION:
The equation given by the intro is 

J = nqv_d (also eq. (30.13) from the book)

The book uses e as the charge, but q is an arbitrary charge, and more applicable to situations not concerning electrons. so q^* will be subbed into other equations that have e. (The asterick demonstrates a substituted variable)

Section 30.2 Creating a Current has a subsection titled: A Model of Conduction. This section is vital to this problem.

eq. (30.7) states that 

v_d = q^*τE/m
and that's actually the answer too

v = Eqτ/m

NOTE: 
At every collision, the electron's motion is randomized, bringing the drift velocity back to zero. If τ is the mean time between collisions, then the speed v that you just calculated is equal to the net drift velocity vd, for all current carriers. This is true, because the mean time between collisions is equal to the mean time since the last collision. (To see how this is possible requires looking at the actual distribution of times between collisions and how the different averages are calculated, but this type of analysis lies beyond the scope of this problem.)
-----------------------------------------------------------------------------------------------------
PART B:
The magnitude of the drift velocity is very small compared to the speed of random electron motion in a metal. The mean time between collisions can be calculated from the mean free path d, which is the average distance that an electron can travel before colliding with one of the metal nuclei. Using these variables, what is the mean time between collisions τ? Let E_F be the energy of the electrons.
Express the mean time between collisions in terms of E_F, d, and m.

SOLUTION:
so... how long, τ, does it take to go a distance, d, at a speed, v_F (the speed of an electron with kinetic energy equal to E_F)?

What equation relates energy velocity and mass? The kinetic energy equation!

K = ½mv²

E_F = ½mv_F²
→ v_F = SQRT{2E_F/m}

velocity is distance traveled over a period of time : v = d/t

therefore, distance traveled is equal to the velocity times time: d = v⋅t
Finally, that means that time elapsed is equal to distance traveled divided by velocity: t = d/v

∴ τ = d/v_F =d⋅ 1/SQRT{2E_F/m}

τ = d⋅SQRT{½m/E_F} = d⋅√{m/(2E_F)}

-----------------------------------------------------------------------------------------------------
PART C:
Recall that the conductivity σ of a substance is defined by the relation J=σE and that the resistivity ρ is the inverse of the conductivity: ρ=1/σ. Give an expression for ρ in terms of microscopic properties of a metal. Use q_e for the charge on the electron.
Express your answer in terms of E_F, m, q_e, d, E, and n.

SOLUTION:
so, if J=σE & ρ=1/σ, then J = E/ρ
eq(30.13) states that J = nev_d

∴ J = E/ρ = nev_d
→ρ = E/[nev_d]

if we plug in the value we got in part A for v, and the instructions statement that e = q_e ... 

ρ = E_F/[nq_e⋅(E_Fq_eτ/m)]

= m/[nq_e²⋅d⋅SQRT{m2E_F}]

If we then plug in the value we got in part B for τ... we should have everything in the proper variables. 

ρ = m/[nq_e²⋅τ]

=m/[nq_e²⋅d⋅SQRT{m/(2E_F)}]
=m⋅SQRT{2E_F/m}/[nq_e²⋅d]
=SQRT{2m²E_F/m}/[nq_e²⋅d]
⇒ ρ = SQRT{2mE_F} / [nq_e²⋅d]

-----------------------------------------------------------------------------------------------------
PART D:
Find the resistivity of gold at room temperature. Use the following information:

• free electron density of gold = 5.90×10²⁸ m⁻³
• Fermi energy of gold = 8.86×10⁻¹⁹ J
• mass of electron = 9.11×10⁻³¹ kg
• charge of an electron = −1.60×10−¹⁹ C
• mean free path of electron in gold = 3.45×10⁻⁸ m

Express your answer in ohm-meters to three significant figures.

SOLUTION:

Just plug these values into the equation we determined in the previous part... 

so what we were given was:

n = 5.90×10²⁸ m⁻³
E_F = 8.86×10⁻¹⁹ J
m = 9.11×10⁻³¹ kg
q_e = −1.60×10−¹⁹ C
d = 3.45×10⁻⁸ m

So ρ = SQRT{2mE_F} / [nq_e²⋅d] 

= SQRT{2(9.11×10⁻³¹ kg)(8.86×10⁻¹⁹ J)} / [(5.90×10²⁸ m⁻³)(−1.60×10−¹⁹ C)²⋅(3.45×10⁻⁸ m)]

⇒ ρ = 2.44×10^-8 Ωm
Wolfram LINK

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