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)

Chapter 25: Overview

CHAPTER 25: ELECTRIC CHARGES & FORCES

In an insulator, the electrons are bound to their molecules... they can shift slightly, creating a rather weak net attraction to a test charge that is brought close to the insulator's surface

In a conductor, free electrons will accumulate on the surface of the conductor nearest the positive test charge. This will create a strong attractive force if the test charge is placed very close to the conductor's surface.

Coulomb’s law:

F = K⋅|q₁|⋅|q₂| / r²

Electric field of a point charge
E = K ⋅ q/r² ⋅ (r^)

Useful:

Electric constant
ε₀ = 8.854×10^-12 F/m

SI Unit conversions
1 N = 1 kg⋅m⋅s^-2
1 C = A⋅s
1 F = A²⋅s⁴⋅kg^-1⋅m^-2

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

25: Magnitude and Direction of Electric Fields

INTRO:
A small object A, electrically charged, creates an electric field. At a point P located 0.250m directly north of A, the field has a value of 40.0N/C directed to the south.

Part A:
What is the charge of object A?

SOLUTION:
givens:
E = 40 N/C 
r = 0.250 m

we can solve for q by rearranging the formula E = K ⋅ q/r²
→ q = E⋅r²/K
q = (40 N/C) ⋅(0.250 m)² ⋅ 4πε₀
→(40 N/C) ⋅ (6.25×10^-2 m²)⋅ 4π(8.854×10^-12 F/m)
q = 2.782×10^-10 C
however, we must also take into consideration the direction of the electric field, since the field is pointing toward the charge, the charge must be negative. Electric fields point away from positive charges
∴ q = -2.782×10^-10 C

Part B:
If a second object B with the same charge as A is placed at 0.250m south of A (so that objects A and B and point P follow a straight line), what is the magnitude of the total electric field produced by the two objects at P?

SOLUTION:
Here we just have to find the net electric field by adding the electric field from A to the electric field from B

E_A = 40 N/C
r_B = r_A + 0.25 m = 0.5 m

E_B = K ⋅ q/r_B²
→[-2.782×10^-10 C] / [4πε₀⋅(0.5 m)²]
→[-2.782×10^-10 C] / [4π(8.854×10^-12 F/m)⋅(0.25 m²)]
E_B = -10 N/C

what this means by -10 N/C is that it's pointing south, just as E_A is
so E_NET = -40 N/C + -10 N/C = -50 N/C 
aka 50 N/C pointing south

25: Charging a Conducting Rod

INTRO:

This problem explores the behavior of charge on conductors. We take as an example a long conducting rod suspended by insulating strings. Assume that the rod is initially electrically neutral. For convenience we will refer to the left end of the rod as end A, and the right end of the rod as end B. In the answer options for this problem, "strongly attracted/repelled" means "attracted/repelled with a force of magnitude similar to that which would exist between two charged balls.

Part A:
A small metal ball is given a negative charge, then brought near (i.e., within about 1/10 the length of the rod) to end A of the rod. What happens to end A of the rod when the ball approaches it closely this first time?

a) It is strongly repelled.
b) It is strongly attracted.
c) It is weakly attracted.
d) It is weakly repelled.
e) It is neither attracted nor repelled.

SOLUTION:
It is stated that the ball is much closer to the end of the rod than the length of the rod. Therefore, if points down the rod several times the distance of approach (but still much closer to end A than end B) are to experience no electric field, the charge on end A of the rod must be comparable in magnitude to the charge on the ball (so that their fields will cancel).

If you can recall the problem from this section, A Test Charge Determines Charge on Insulating and Conducting Balls, and the points made regarding conductors, it can be ascertained that in conductors, the electrons are free to move about. This means that when a charge is brought near to a conductor, the opposite charges all navigate to the point closest the charge and a strong attraction is created
this is true for a conducting rod as well, so the correct answer is option b)

NOTE:
Now consider what happens when the small metal ball is repeatedly given a negative charge and then brought into contact with end A of the rod.

Part B:
After a great many contacts with the charged ball, how is the charge on the rod arranged (when the charged ball is far away)?

a) There is positive charge on end B and negative charge on end A.
b) There is negative charge spread evenly on both ends.
c) There is negative charge on end A with end B remaining neutral.
d) There is positive charge on end A with end B remaining neutral.

SOLUTION: 
in a conductor, no matter how many times it comes in contact with a charge, if it returns to a static situation the charge returns to being neutral. This is satisfied by option b)

Part C:
How does end A of the rod react when the charged ball approaches it after a great many previous contacts with end A? Assume that the phrase "a great many" means that the total charge on the rod dominates any charge movement induced by the near presence of the charged ball.

a) It is strongly repelled.
b) It is strongly attracted.
c) It is weakly attracted.
d) It is weakly repelled.
e) It is neither attracted nor repelled.

SOLUTION:

The answer is a) 
but I can't exactly explain why
I am rather confused from this one 

Part D:
How does end B of the rod react when the charged ball approaches it after a great many previous contacts with end A?

a) It is strongly repelled.
b) It is strongly attracted.
c) It is weakly attracted.
d) It is weakly repelled.
e) It is neither attracted nor repelled.

SOLUTION:
Because the rod is a conductor, the charge is free to distribute itself over the entire rod. It must be distributed so that the internal electric field in the rod is zero, and there is only one distribution that achieves this. There is no memory in this situation: The charge will always distribute itself into the same final result. The rod is symmetric. Therefore, the final distribution of charge must also be symmetric, and hence the same charge must be on end A as on end B.
so the answer is a)