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.)

26: PhET Tutorial: Charges and Electric Fields

Part A: 
The electric field produced by the positive charge

a) is directed radially toward the charge at all locations near the charge.
b) wraps circularly around the positive charge.
c) is directed radially away from the charge at all locations near the charge.

SOLUTION:
This means that another positive charge, if placed near the original charge, would experience a force directed radially away from the original charge.

Part B:
For these four locations, the magnitude of the electric field is

a) greatest to the left of the charge.
b) the same.
c) greatest below the charge.
d) greatest to the right of the charge.
e) greatest above the charge.

SOLUTION:
This result implies that the strength of the electric field due to one point charge depends solely on the distance away from the charge. Mathematically, we say the electric field is spherically symmetric.

Part C:
The magnitude of the electric field 1 m away from the positive charge is 
a) one-half
b) equal to
c) two times
d) four times
e) one-quarter
the magnitude of the electric field 2 m away.

SOLUTION:
The magnitude of the field decreases more quickly than the inverse of the distance from the charge. The magnitude of the electric field is proportional to the inverse of the distance squared (E∝1/r², where r is the distance from the charge). You should verify this by looking at the field strength 3 or 4 meters away. This is consistent with Coulomb'€™s law, which states that the magnitude of the force between two charged particles is F=kQ₁Q₂/r².

Part D:
If the field strength is E = 9 V/m a distance of 1 m from the charge, what is the field strength E a distance of 3 m from the charge?

SOLUTION:
E∝1/r², so E₂ = 9/3² = 1 V/m

Part E:
Which statement best describes the differences in the electric field due to a negative charge as compared to a positive charge?

a) The electric field changes direction (now points radially inward), but the electric field strength does not change.
b) Nothing changes; the electric field remains directed radially outward, and the electric field strength doesn'™t change.
c) The electric field changes direction (now points radially inward), and the magnitude of the electric field decreases at all locations.

SOLUTION:
The electric field is now directed toward the negative charge, but the field strength doesn'€™t change. The electric field of a point charge is given by E⃗ =(kQ/r²)r^. Because of the sign of the charge, the field produced by a negative charge is directed opposite to that of a positive charge but the magnitude of the field is the same.

Part F:
Where is the magnitude of the electric field roughly equal to zero (other than very far away from the charges)?

a) The electric field is nonzero everywhere on the screen.
b) The electric field is roughly zero near the midpoint of the two charges.
c) The electric field is zero at any location along a vertical line going through the point directly between the two charges.

SOLUTION: 
Directly between the two charges, the electric fields produced by each charge are equal in magnitude and point in opposite directions, so the two vectors add up to zero.

Part G:
Consider a point 0.5 m above the midpoint of the two charges. As you can verify by removing one of the positive charges, the electric field due to only one of the positive charges is about 18 V/m. What is the magnitude of the total electric field due to both charges at this location?

a) 36 V/m
b) 25 V/m
c) zero

SOLUTION:
Notice that this number is less than twice the magnitude of the field due to each charge. This occurs because the horizontal components of the electric field due to each charge exactly cancel out (add to zero). Only the vertical components of the fields add together.

Part H:
The electric field at the midpoint is

a) directed to the left
b) zero
c) directed to the right

SOLUTION:
The electric field due to the positive charge is directed to the right, as is the electric field due to the negative charge. So the net electric field, which is the sum of these two fields, is also to the right.

Part I:
Measure the strength of the electric field 0.5 m directly above the midpoint as well as 1 m directly above. Does the strength of the electric field decrease as 1 over distance squared (1/r²)?

a) No, it decreases less quickly with distance
b) No, it decreases more quickly with distance
c) Yes, it does

SOLUTION:
In fact, it turns out that the strength of the electric field decreases roughly as 1/r³! So the field 1 m above the midpoint is roughly eight times weaker than at 0.5 m above the midpoint. The important lesson here is that, in general, a distribution of charges produces an electric field that is very different from that of a single charge.

Part J:
Measure the strength of the electric field 1 m directly above the middle as well as 2 m directly above. Does the strength of the electric field decrease as 1 over distance squared (1/r²)?

a) No, it decreases less quickly with distance
b) No, it decreases more quickly with distance
c) Yes, it does

SOLUTION:
In fact, it turns out that the strength of the electric field decreases roughly as 1/r. So the field 1 m above the midpoint is roughly half the strength at 0.5 m. This is another example showing that a distribution of charges produces an electric field that is very different from that of a single charge.

26: Electric Field due to Multiple Point Charges

INTRO:
Two point charges are placed on the x axis. The first charge, q₁ = 8.00 nC , is placed a distance 16.0 m from the origin along the positive x axis; the second charge, q₂ = 6.00 nC , is placed a distance 9.00 m from the origin along the negative x axis.

Part A: 
Calculate the electric field at point A, located at coordinates (0 m, 12.0m).
Give the x and y components of the electric field as an ordered pair. Express your answer in newtons per coulomb to three significant figures.

SOLUTION:
Find the contributions to the electric field at point A separately for q₁ and q₂, then add them together using vector addition to find the total electric field at that point. You will need to use the Pythagorean theorem to find the distance of each charge from point A.

Triangle 1 ⇒ ΔOAq₁
a = 12 m; b = 16 m
∴c = 20 m
Triangle 2 ⇒ ΔOAq₂
a = 12 m; b = 9 m
∴c = 15 m

Now find the contributions to the electric field at point A
E_A1 = K⋅q₁/r² = K⋅(8.00 nC)/(20 m)² = 0.1798 N/C & points in direction <-16, 12>
E_A2 = K⋅q₂/r² = K⋅(6.00 nC)/(15 m)² = 0.2397 N/C & points in direction <9, 12>

Now calculate the x and y components of both electric fields 
to get the angle for E_A1, θ = tan^-1(12/16) = 36.87°
E_A1x = E_A1 cos(θ) = 0.1798 N/C⋅0.8 = 0.14384 N/C
E_A1y = E_A1 sin(θ) = 0.1798 N/C⋅0.6 = 0.10788 N/C
E_A1 = <-0.14384, 0.10788> N/C
to get the angle for E_A2, θ = tan^-1(12/9) = 53.13°
E_A2x = E_A2 cos(θ) = 0.2397 N/C⋅0.6 = 0.14382 N/C
E_A2y = E_A2 sin(θ) = 0.2397 N/C⋅0.8 = 0.19176 N/C 
E_A2 = <0.14382, 0.19176> N/C

SO, now that you have the vector components of the two electric fields, you are able to use vector addition to determine the net electric field at point A
E_NET = E_A1 + E_A2 = <-0.144, 0.108> N/C + <0.144 0.192> N/C = <0, 0.300> N/C

Part B:
An unknown additional charge q₃ is now placed at point B, located at coordinates (0 m, 15.0m).

Find the magnitude and sign of q₃ needed to make the total electric field at point A equal to zero.
Express your answer in nanocoulombs to three significant figures.

SOLUTION:
luckily q₃ is in line with point A so we don't need to worry about x and y components as with the previous part. 
givens:
r_AB = 3 m
E_AB = <0, 0.300> N/C

E_AB = K q₃/r_AB²
q₃ = E_AB⋅r_AB² / K
→0.300 N/C ⋅ (3 m)² / [4πε₀]
q₃ = 3×10^-10 C = 0.300 nC

25: Video Tutor: Charged Rod and Aluminum Can

Part A: 
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?

a) Rolls to the left
b) Stays still
c) Rolls to the right 

SOLUTION:
The positively charged rod induces a negative charge on the left side of the can, creating an attractive force between the rod and the can. However, the negatively charged rod induces an equal positive charge on the right side of the can, which creates an attractive force between the can and that rod. The net force acting on the can is zero.
So the answer is option b)

Part B:
Now, consider the situation shown in the figure below. What does the can do?

a) Stays still
b) Rolls to the right 
c) Rolls to the left

SOLUTION:
The polarization force is always attractive, so the can does not move, option a)

Part C:
Using the setup from the first question, imagine that you briefly touch the negatively charged rod to the can. You then hold the two rods at equal distances on either side of the can. What does the can do?

a) Does not move
b) Rolls toward the positively charged rod
c) Rolls away from the positively charged rod 

SOLUTION:
The can acquires a net negative charge after being touched, so it is then attracted to the positively charged rod, or option b)

25: Problem 25.17

INTRO:

Part A:
What is the magnitude of the net electric force on charge B in the figure?
Assume a = 2.0 cm and b = 1.3 cm .
Express your answer to two significant figures and include the appropriate units.

SOLUTION:
F_AonB= K⋅|q_A⋅q_B| / [a]² = 1/[4πϵ₀]⋅|(-1.0 nC)⋅(-2.0 nC)| / [2.0 cm]² = 4.494×10^-5N
F_ConB= K⋅|q_C⋅q_B| / ²= 1/[4πϵ₀]⋅|(2.0 nC)⋅(-2.0 nC)| / [1.3 cm]² = 2.13×10^-4N

F_NETonB = F_AonB + F_ConB = 4.494×10^-5N + 2.13×10^-4N = 2.579×10^-4N ≈ 2.6×10^-4 N

Part B:
What is the direction of the net electric force on charge B in the figure?

SOLUTION:
In the previous part, we determined the magnitude of the net force on charge B
However we must also determine the direction of the force

⇐((-)A) ((+)B) ⇐((+)C)
F_AonB is a negative charge pushing the negative charge down
F_ConB is a positive charge pulling the negative charge down

SO,
the force is directed down

25: Forces in a Three-Charge System

INTRO:
Coulomb's law for the magnitude of the force F between two particles with charges Q and Q′ separated by a distance d is |F|=K⋅|QQ′| / d²

where K=1/[4πϵ₀], and ϵ₀=8.854×10⁻¹²C²/(N⋅m²) is the permittivity of free space.

Consider two point charges located on the x axis: one charge, q₁ = -19.5 nC , is located at x₁ = -1.670 m ; the second charge, q₂ = 34.0 nC , is at the origin (x=0).

Part A:
What is the net force exerted by these two charges on a third charge q₃ = 52.5 nC placed between q₁ and q₂ at x₃ = -1.075 m ?

Your answer may be positive or negative, depending on the direction of the force.
Express your answer numerically in newtons to three significant figures.

SOLUTION:
givens:
q₁ = -19.5 nC
x₁ = -1.670 m
q₂ = 34.0 nC
x₂ = 0 m
q₃ = 52.5 nC
x₃ = -1.075 m

we begin by determining the two separate forces on q₃
F_1on3= K⋅|q₁⋅q₃| / [x₃-x₁]² = 1/[4πϵ₀]⋅|(-19.5 nC)⋅(52.5 nC)| / [(-1.075 m)-(-1.670 m)]² = 2.6×10^-5 N
F_2on3= K⋅|q₂⋅q₃| / [x₃-x₂]²= 1/[4πϵ₀]⋅|(34.0 nC)⋅(52.5 nC)| / [(-1.075 m)-(0 m)]² = 1.388×10^-5 N

however these are only the magnitudes of the forces, we also need to determine the directions so as to assign the forces positive or negative values
in other words, is each force pushing the charge in a positive x-direction or a negative x-direction?
⇐((-)q1) ((+)q3) ⇐((+)q2) 
F_1on3 is a negative charge pulling the positive charge in the negative x-direction
F_2on3 is a positive charge pushing the positive charge in the negative x-direction

SO, F_NETon3 = F_1on3(negative) + F_2on3(negative) =-2.6×10^-5 N + -1.388×10^-5 N = -3.99×10^-5 N