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.
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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... 
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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)

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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: 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.
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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
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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
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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

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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:

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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
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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

30: Problem 30.51

INTRO:
A hollow metal sphere has inner radius a, outer radius b, and conductivity σ. The current I is radially outward from the inner surface to the outer surface.
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PART A:
Find an expression for the electric field strength inside the metal as a function of the radius r from the center.
Express your answer in terms of the variables I, σ, r, and appropriate constants.

SOLUTION:
If you were to consider a thin radial section of the cylinder (so that the cross section would be almost uniform throughout), what would be its resistance? 

The radial area, not cross-sectional this time, is A = 2πrL
the L = 2r ∴ A = 4πr²

J = I/A = σE 

→ E = I/(σA) = I/(σ4πr²)

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PART B:
Evaluate the electric field strength at the inner surface of a copper sphere if a = 1.0 cm , b = 3.0 cm , and I = 26 A .
Express your answer to two significant figures and include the appropriate units.

SOLUTION:

Givens/ Conversions:

a = 1.0 cm = 1×10^-2 m
b = 3.0 cm = 3×10^-2 m
I = 26 A
material = copper,

table (30.2) → σ = 6.0×10⁷ 1/(Ωm)

they want the strength at the inner surface, so use a as r...

→ E = (26 A)/[4π(6.0×10⁷ 1/(Ωm))(1×10^-2 m)²]
⇒ E = 3.45×10^-4 V/m

They want E = 3.4×10^-4 V/m though... for some reason.  It'll automatically correct it though.

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PART C:
Evaluate the electric field strength at the outer surface of a copper sphere if a = 1.0 cm , b = 3.0 cm , and I = 26 A .
Express your answer to two significant figures and include the appropriate units.

SOLUTION:

Givens/ Conversions:

a = 1.0 cm = 1×10^-2 m
b = 3.0 cm = 3×10^-2 m
I = 26 A
material = copper,

table (30.2) → σ = 6.0×10⁷ 1/(Ωm)

they want the strength at the inner surface, so use a as r...

→ E = (26 A)/[4π(6.0×10⁷ 1/(Ωm))(3×10^-2 m)²]
⇒ E = 3.83×10^-4 V/m

They want E = 3.8×10^-4 V/m ... At least that rounding makes sense :) 

30: Problem 30.24

INTRO:
The two segments of the wire in the figure (Figure 1) have equal diameters but different conductivities σ₁ and σ₂. Current I passes through this wire.

Figure 1

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PART A:
If the conductivities have the ratio σ₂/σ₁=5, what is the ratio E₂/E₁ of the electric field strengths in the two segments of the wire?

- 1/25
- 5
- 25
- 1/5

SOLUTION:
Since the current I passes through the entire wire, and the diameters are equal... the current density for both sides is... 
J = I/A

This means that J₁ = J₂

using eq. (30.17) ... J = σE ...

J₁ = σ₁E₁
J₂ = σ₂E₂​

since J₁ = J₂... σ₁E₁ = σ₂E₂

by rearranging...
σ₂/σ₁ = E₁/E₂ = 5 (given)​

E₂/E₁ = [E₁/E₂]^-1

⇒ E₂/E₁ = 1/5​

30: Problem 30.19

INTRO:
The electric field in a 1.0mm×1.0mm square aluminum wire is 2.8×10⁻² V/m .
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PART A:
What is the current in the wire?
Express your answer to two significant figures and include the appropriate units.

SOLUTION:
E = 2.8×10⁻² V/m
b = h = 1.0 mm = 0.001 m

∴ A = 1×10^-6 m²​

using table 30.2... 

σ_Al = 3.5×10⁷ 1/(Ωm)​

Using eq. (30.17) & eq. (30.13)... 
J = I/A = σE

→ I = σEA

= (3.5×10⁷ 1/(Ωm))(2.8×10⁻² V/m)(1×10^-6 m²)
⇒ I = 0.98 A

* note 1 V/Ω = A ( since V=IR ) *​

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 .
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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

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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​

28: Problem 28.2

INTRO:
The electric field strength is 2.50×10⁴ N/C inside a parallel-plate capacitor with a 1.50 mm spacing. An electron is released from rest at the negative plate.

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PART A:
What is the electron's speed when it reaches the positive plate?
Express your answer with the appropriate units.

SOLUTION:
Givens / conversions:
E = 2.50×10⁴ N/C 
d =1.50 mm = 0.0015 m

we also need to know the properties of an electron
q_e = -1.60×10^-19 C
m_e = 9.11×10^-31 kg

Using the energy equations:
ΔPE = U_f - U_i = qEd₂ - qEd₁
since it goes from x = d to x = 0,

ΔPE = qEd​

ΔKE = K_f - K_i = ½m(v_f² - v_i²)
since we know that it starts at rest... 

ΔKE = ½mv_f²​

ΣEnergy = 0: ΔPE + ΔKE = 0
qEd + ½mv_f² = 0

now solve for v_f... 
v_f = [-2qEd/m]^½

plug in values... 
v_f = [-2(-1.60×10^-19 C)⋅(2.50×10⁴ N/C )⋅(0.0015 m)/(9.11×10^-31kg)]^½

⇒ ∴ v_f =​ 3.629×10⁶ m/s

v_f =​ 3.63×10⁶ m/s
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28: PhET Tutorial: Charges and Electric Potential

INTRO:
Learning Goal:
To understand the spatial distribution of the electric potential for a variety of simple charge configurations, and to understand how the electric field and electric potential (voltage) are related.
For this problem, use the PhET simulation Charges and Fields. This simulation allows you to place multiple positive and negative point-charges in any configuration and look at the resulting electric field and corresponding electric potential.

[ SIMULATION ]

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PART A:
The electric potential (voltage) at a specific location is equal to the potential energy per unit charge a charged object would have if it were at that location. If the zero point of the voltage is at infinity, the numerical value of the voltage is equal to the numerical value of work done to bring in a unit charge from infinity to that location.
Select Show numbers and grid in the green menu, and drag one positive charge to the middle of the screen, right on top of two intersecting bold grid lines.
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?


Express your answer numerically in volts to two significant figures.

SOLUTION:
Unlike the magnitude of the electric field, the electric potential (voltage) is not proportional to the inverse of the distance squared.

4.5 V

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PART B:
What is the voltage 3 m away from the charge?

- 3 V
- 9 V
- 1 V

SOLUTION:
3 V

NOTE: 
Based on this result, and the previous question, the electric potential (voltage) is inversely proportional to the distance r from the charge: V∝1/r. Recall that the magnitude of the electric field E∝1/r².
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PART C:
Another way to study voltage and its relationship to electric field is by producing equipotential lines. Just like every point on a contour line has the same elevation in a topographical map, every point on an equipotential line has the same voltage.
Click plot on the voltage tool to produce an equipotential line. Produce many equipotential lines by clicking plot as you move the tool around.  You should produce a graph that looks similar to the one shown below.

Place several E-Field Sensors at a few points on different equipotential lines, and look at the relationship between the electric field and the equipotential lines. Which statement is true?

- At any point, the electric field is parallel to the equipotential line at that point.
- At any point, the electric field is perpendicular to the equipotential line at that point, and it is directed toward lines of lower voltages.
- At any point, the electric field is perpendicular to the equipotential line at that point, and it is directed toward lines of higher voltages.

SOLUTION:
The second option is true,
At any point, the electric field is perpendicular to the equipotential line at that point, and it is directed toward lines of lower voltages.

NOTE:
All points on an equipotential line have the same voltage; thus, no work would be done in moving a test charge along an equipotential line. No work is done because the electric field, and thus the force on the test charge, is perpendicular to the displacement of the test charge being moved along the equipotential line.
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PART D:
Equipotential lines are usually shown in a manner similar to topographical contour lines, in which the difference in the value of consecutive lines is constant. Clear the equipotential lines using the Clear button on the voltage tool. Place the first equipotential line 1 m away from the charge. It should have a value of roughly 9 V. Now, produce several additional equipotential lines, increasing and decreasing by an interval of 3 V (e.g., one  with 12 V, one with 15 V, and one with 6 V). Don’t worry about getting these exact values. You can be off by a few tenths of a volt.Which statement best describes the distribution of the equipotential lines?

- The equipotential lines are closer together in regions where the electric field is stronger.
- The equipotential lines are closer together in regions where the electric field is weaker.
- The equipotential lines are equally spaced. The distance between each line is the same for all adjacent lines.

SOLUTION:
The first option is the correct one

The equipotential lines are equally spaced. The distance between each line is the same for all adjacent lines.

*UPDATE 07/26/2017 @ Andrea S.*
If you look closely, you can see that the first option is ACTUALLY: 
The equipotential lines are closer together in regions where the electric field is stronger


NOTE:

Near the positive charge, where the electric field is strong, the voltage lines are close to each other.  Farther from the charge, the electric field is weaker and the lines are farther apart.

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PART E:

Now, remove the positive charge by dragging it back to the basket, and drag one negative charge toward the middle of the screen. Determine how the voltage is different from that of the positive charge.How does the voltage differ from that of the positive charge?

- The voltages are positive, but the magnitude increases with increasing distance.

- The voltage distribution does not change.

- The voltages become negative instead of positive and keep the same magnitudes.

SOLUTION:

The third option is true

The voltages become negative instead of positive and keep the same magnitudes.

NOTE:

The voltage is still inversely proportional to the distance from the charge, but the voltage is negative everywhere rather than positive.

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PART F:

Now, remove the negative charge, and drag two positive charges, placing them 1 m apart, as shown below.

What is the voltage at the midpoint of the two charges?

- Exactly twice the voltage produced by only one of the charges at the same point

- Zero

- Greater than zero, but less than twice the voltage produced by only one of the charges at the same point

SOLUTION:

The first one is correct

Exactly twice the voltage produced by only one of the charges at the same point

NOTE:

Because voltage is a scalar quantity, there are no vector components with opposite directions canceling out, as for electric fields. The voltage is simply the sum of the voltages due to each of the individual charges. Since both charges are positive, the voltage due to each charge (at all locations) is positive.

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PART G:

Now, make an electric dipole by replacing one of the positive charges with a  negative charge, so the final configuration looks like the figure shown below.

What is the voltage at the midpoint of the dipole?

The voltage at the midpoint of the dipole is...

- positive
- zero
- negative

SOLUTION:

The voltage at the midpoint is zero

NOTE:

Because the voltage due to the negative charge has the opposite sign of the voltage due to the positive charge at the midpoint, the net voltage is zero. The electric field, however, is not zero here!-----------------------------------------------------------------------------------------------------

PART H:

Make several equipotential lines similar to the figure below.

Try to have the equipotential lines equally spaced in voltage. Then, use an E-Field Sensor to measure the electric field at a few points while looking at the relationship between the electric field and the equipotential lines.
Which of the following statements is true?

- The electric field strength is greatest where the equipotential lines are very close to each other.
- The electric field strength is greatest where the voltage is the smallest.
- The electric field strength is greatest where the voltage is the greatest.

SOLUTION:

The first option is correct
The electric field strength is greatest where the equipotential lines are very close to each other.

HINT:

Locations where the voltage is changing steeply are locations with a strong electric field. The magnitude of the electric field is equal to the rate the voltage is changing with distance. Mathematically, this idea is conveyed by |Es|=dV/ds, where Es is the component of the electric field in the direction of a small displacement ds. (As you learned earlier, the electric field is directed in the direction where the voltage decreases.)