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.)

27: The Electric Field of a Ball of Uniform Charge Density

INTRO:
A solid ball of radius r_b has a uniform charge density ρ.
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PART A:
What is the magnitude of the electric field E(r) at a distance r>r_b from the center of the ball?
Express your answer in terms of ρ, r_b, r, and ϵ₀.

SOLUTION:
Since a charge density is given, the charge must be acquired from this density.
A density has the units [something] per volume. Therefore, the charge can be determined if the volume of the sphere is determined first. 
V = (4/3)πr³
∴ q = ρ⋅V = ρ⋅(4/3)πr³

E(r) = ρr_b³ / (3ε₀⋅r²)

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PART B:
What is the magnitude of the electric field E(r) at a distance r<r_b from the center of the ball?
Express your answer in terms of ρ, r_b, r, and ϵ₀.

SOLUTION:
Since the Gaussian surface is inside the ball, the surface encloses only a fraction of the ball's charge. This is why Gauss's law is so useful: As long as it is symmetrically distributed, the charge outside the Gaussian surface is irrelevant in calculating the field!

The net charge enclosed by the Gaussian surface is different than that of the net charge enclosed in part A.

E(r) = ρr / (3ε₀)

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PART C:

Let E(r) represent the electric field due to the charged ball throughout all of space. Which of the following statements about the electric field are true?
Check all that apply.

• E(0) = 0
• E(r_b)=0
• lim_r→∞E(r)=0
• The maximum electric field occurs when r=0
• The maximum electric field occurs when r=r_b
• The maximum electric field occurs as r→∞.

SOLUTION:

E(0) = 0
lim_r→∞E(r)=0
The maximum electric field occurs when r=r_b

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27: Problem 27.47

INTRO:
The three parallel planes of charge shown in the figure (Figure 1) have surface charge densities −1/2 η, η, and −1/2 η.

Figure 1

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PART A:
Find the magnitude of the electric field in region 1.

SOLUTION:
<< explanation to be added >>
0 η/ε₀

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PART B:

What is the direction of the electric field in region 1?

• Upward
• Downward
• The field is zero

SOLUTION:
<< explanation to be added >>
The field is zero

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PART C:

Find the magnitude of the electric field in region 2.

SOLUTION:

<< explanation to be added >>

0.500 η/ε₀

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PART D:

What is the direction of the electric field in region 2?

• Upward
• Downward
• The field is zero

SOLUTION:

<< explanation to be added >>

Upward

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PART E:

Find the magnitude of the electric field in region 3.

SOLUTION:

<< explanation to be added >>

0.500 η/ε₀

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PART F:

What is the direction of the electric field in region 3?

• Upward
• Downward
• The field is zero

SOLUTION:

<< explanation to be added >>

Downward

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PART G:

Find the magnitude of the electric field in region 4.

SOLUTION:

<< explanation to be added >>

0 η/ε₀

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PART H:

What is the direction of the electric field in region 4?

• Upward
• Downward
• The field is zero

SOLUTION:

<< explanation to be added >>

The field is zero

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27: Problem 27.46

INTRO:
A uniformly charged ball of radius a and charge −Q is at the center of a hollow metal shell with inner radius b and outer radius c. The hollow sphere has net charge +2Q.


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PART A: 
Determine the magnitude of the electric field in the region r≤a. Give your answer as a multiple of Q/ε₀.
Express your answer in terms of some or all of the variables a, b, c, r, and the constant π.

SOLUTION:
A_sphere = 4πr²

Φ_E = ∫E⋅dA = q/ε₀
E = q/(A⋅ε₀) = q/(4πr²ε₀)

for r≤a, q_in = -Q
E = -r/(4πa³)⋅Q/ε₀ r^

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PART B:
Determine the magnitude of the electric field in the region a<r<b. Give your answer as a multiple of Q/ε₀.
Express your answer in terms of some or all of the variables a, b, c, r, and the constant π.

SOLUTION:
<< explanation to be added >>
E = -1/(4πr²)⋅Q/ε₀ r^

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PART C:

Determine the magnitude of the electric field in the region b<r<c. Give your answer as a multiple of Q/ε₀.

Express your answer in terms of some or all of the variables a, b, c, r, and the constant π.

SOLUTION:

<< explanation to be added >>

0

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PART D:

Determine the magnitude of the electric field in the region c<r. Give your answer as a multiple of Q/ε₀.

Express your answer in terms of some or all of the variables a, b, c, r, and the constant π.

SOLUTION:

<< explanation to be added >>

E = 1/(4πr²)⋅Q/ε₀ r^

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