An electric field E=100,000 ı^ N/C causes the 5.0 g point charge in the figure to hang at a 20° angle.
 |
| Figure 1 |
What is the charge on the ball?
Express your answer to two significant figures and include the appropriate units.
SOLUTION:
q = 180 nC
Showing posts with label Chapter 25. Show all posts
Showing posts with label Chapter 25. Show all posts
Wednesday, September 30, 2015
25: Problem 25.66
INTRO:
An electric field E=100,000 ı^ N/C causes the 5.0 g point charge in the figure to hang at a 20° angle.
Part A:
What is the charge on the ball?
Express your answer to two significant figures and include the appropriate units.
SOLUTION:
q = 180 nC
An electric field E=100,000 ı^ N/C causes the 5.0 g point charge in the figure to hang at a 20° angle.
What is the charge on the ball?
Express your answer to two significant figures and include the appropriate units.
SOLUTION:
q = 180 nC
25: Problem 25.59
INTRO:
Two 3.40g point charges on 1.00-m-long threads repel each other after being equally charged, as shown in the figure.Assume that θ = 28
Part A:
What is the charge q?
Express your answer to three significant figures and include the appropriate units.
SOLUTION:
q = -1.32 μC
25: Problem 25.57
INTRO:
You sometimes create a spark when you touch a doorknob after shuffling your feet on a carpet. Why? The air always has a few free electrons that have been kicked out of atoms by cosmic rays. If an electric field is present, a free electron is accelerated until it collides with an air molecule. It will transfer its kinetic energy to the molecule, then accelerate, then collide, then accelerate, collide, and so on. If the electron’s kinetic energy just before a collision is 1.80×10−18J or more, it has sufficient energy to kick an electron out of the molecule it hits. Where there was one free electron, now there are two! Each of these can then accelerate, hit a molecule, and kick out another electron. Then there will be four free electrons. In other words, as shows below, a sufficiently strong electric field causes a “chain reaction” of electron production. This is called a breakdown of the air. The current of moving electrons is what gives you the shock, and a spark is generated when the electrons recombine with the positive ions and give off excess energy as a burst of light.
Part A:
The average distance an electron travels between collisions is 2.00 μm . What acceleration must an electron have to gain 1.80×10−18J of kinetic energy in this distance?
Express your answer to three significant figures and include the appropriate units.
SOLUTION:
a = 9.88×1017 m/s2
Part B:
What force must act on an electron to give it the acceleration found in part A?
Express your answer to three significant figures and include the appropriate units.
SOLUTION:
F = 9.00×10-13 N
Part C:
What strength electric field will exert this much force on an electron? This is the breakdown field strength.
Note: The measured breakdown field strength is a little less than your calculated value because our model of the process is a bit too simple. Even so, your calculated value is close.
Express your answer to three significant figures and include the appropriate units.
SOLUTION:
E = 5.63×106 N/C
Part D:
Suppose a free electron in air is 0.700 cm away from a point charge. What minimum charge must this point charge have to cause a breakdown of the air and create a spark? Assume the electron is not hitting the charge particle but a neutral air atom that is still 2.00 μm away.
Express your answer to three significant figures and include the appropriate units.
SOLUTION:
q = 30.6 nC
25: Problem 25.51
INTRO:
In a simple model of the hydrogen atom, the electron moves in a circular orbit of radius 0.053 nm around a stationary proton.
Part A:
How many revolutions per second does the electron make?
Express your answer to two significant figures and include the appropriate units.
SOLUTION:
6.6×1015 rev/s
UPDATE, 09/12/2016
The first thing I always do is convert my givens to SI units
Now. We know an ELECTRON is moving around a PROTON, so we can look up the charges for these two items
Next we need to do a basic dynamics problem.
Since we're trying to find the velocity and know everything else except the force (and this entire section has to do with Force & Coulomb's Law...)
v = (ke⋅q1⋅q2 / (r⋅m) )½
In a simple model of the hydrogen atom, the electron moves in a circular orbit of radius 0.053 nm around a stationary proton.
Part A:
How many revolutions per second does the electron make?
Express your answer to two significant figures and include the appropriate units.
SOLUTION:
6.6×1015 rev/s
UPDATE, 09/12/2016
The first thing I always do is convert my givens to SI units
r = 0.053 nm = 5.3×10-2 nm = 5.3×10-11 m
Now. We know an ELECTRON is moving around a PROTON, so we can look up the charges for these two items
q1 = qPROTON = 1.602×10-19 C
q2 = qELECTRON = -1.602×10-19 C
q2 = qELECTRON = -1.602×10-19 C
Next we need to do a basic dynamics problem.
There is an electromagnetic force pulling the electron around the neutron. F = ma
the acceleration is tangential to the orbit, so a = an = ω2⋅r
ω = v/r ∴ ω2 = v2 / r2 ∴ an = v2 / r2 ⋅ r = v2 / r
⇒ F = m⋅an = m⋅v2 / r
the acceleration is tangential to the orbit, so a = an = ω2⋅r
ω = v/r ∴ ω2 = v2 / r2 ∴ an = v2 / r2 ⋅ r = v2 / r
⇒ F = m⋅an = m⋅v2 / r
Since we're trying to find the velocity and know everything else except the force (and this entire section has to do with Force & Coulomb's Law...)
F = ke⋅q1⋅q2/r2
∴ ke⋅q1⋅q2/r2 = m⋅v2 / r
⇒ ke⋅q1⋅q2 / (r⋅m) = v2
∴ v = (ke⋅q1⋅q2 / (r⋅m) )½
The mass of an electron is me = 9.109×10-31 kg∴ ke⋅q1⋅q2/r2 = m⋅v2 / r
⇒ ke⋅q1⋅q2 / (r⋅m) = v2
∴ v = (ke⋅q1⋅q2 / (r⋅m) )½
v = (ke⋅q1⋅q2 / (r⋅m) )½
= SQRT{ (8.99×109 Nm2/C2) ⋅ (1.602×10-19 C) ⋅ (-1.602×10-19 C) ⋅ 1/(9.109×10-31 kg) ⋅ 1/(5.3×10-11 m) ⋅ (kg⋅m/s2)/N }
= SQRT{ -4.779 × 1012 m2 / s2 }
--> v = 2.186 × 106 m/s
convert to rev/s -> 1 rev / (2 π r) = 1 rev / 3.33×10-10 m
v = 6.565 rev/s
*update (07/26/2017 @ Kirtan Jani) * v = 6.565×1015 rev/s = 6.6×1015 rev/s
This was wrong because I missed an exponent while explaining. However, if you look at the purple highlighted answer above.... you can see that that was only wrong because you were being too GD judgemental. Remember this is a free service bruh..
25: Problem 25.40
Part A:
What is the force F⃗ on the 1 nC charge at the bottom?
Write your answer as two vector components, separated by a comma. Express each component numerically, in newtons, to two significant figures.
SOLUTION:
Fx, Fy = 0, 1.1×10-5 N
25: Problem 25.33
Part A:
What is the magnitude of the force F⃗ on the 1.0 nC charge in the figure?
Express your answer to two significant figures and include the appropriate units.
SOLUTION:
F = 3.1×10-4 N
Here's why ->
Part B:
What is the direction of the force F⃗ on the 1.0 nC charge in the figure?
SOLUTION:
Upward
What is the magnitude of the force F⃗ on the 1.0 nC charge in the figure?
Express your answer to two significant figures and include the appropriate units.
SOLUTION:
F = 3.1×10-4 N
Here's why ->
Part B:
What is the direction of the force F⃗ on the 1.0 nC charge in the figure?
SOLUTION:
Upward
25: Problem 25.9
INTRO:
Two neutral metal spheres on wood stands are touching. A negatively charged rod is held directly above the top of the left sphere, not quite touching it. While the rod is there, the right sphere is moved so that the spheres no longer touch. Then the rod is withdrawn.
Part A:
Afterward, what is the charge state of each sphere?
a) Both the spheres are neutral.
b) The left sphere is negatively charged, the right sphere is charged positively.
c) The right sphere is negatively charged, the left sphere is charged positively.
d) Both the spheres are charged positively.
e) Both the spheres are charged negatively.
SOLUTION:
option c)
Two neutral metal spheres on wood stands are touching. A negatively charged rod is held directly above the top of the left sphere, not quite touching it. While the rod is there, the right sphere is moved so that the spheres no longer touch. Then the rod is withdrawn.
Part A:
Afterward, what is the charge state of each sphere?
a) Both the spheres are neutral.
b) The left sphere is negatively charged, the right sphere is charged positively.
c) The right sphere is negatively charged, the left sphere is charged positively.
d) Both the spheres are charged positively.
e) Both the spheres are charged negatively.
SOLUTION:
option c)
25: Problem 25.30
INTRO:
The nucleus of a 125Xe atom (an isotope of the element xenon with mass 125 u) is 6.0 fm in diameter. It has 54 protons and charge q = +54e.
Part A:
What is the electric force on a proton 1.4 fm from the surface of the nucleus?
Hint: Treat the spherical nucleus as a point charge.
Express your answer to two significant figures and include the appropriate units.
SOLUTION:
We know the formula for the force between two charged particles is F = K |q1| |q2| r-2
We simply have to determine which values to plug into this equation and we will get the answer that we're looking for.
Begin by taking the radius of the nucleus, which is half the diameter, or 3 fm
the added distance of the proton is 1.4 fm, so the total distance of the proton from the center of the nucleus is 3 + 1.4 = 4.4 fm = 4.4⋅10-15 m
We're given that the nucleus has a charge of q = +54e.
Knowing that e=1.6⋅10-19 C , qnucleus = 8.64⋅10-18 C
(remember K is the electrostatic constant and equal to 9.0⋅109 Nm2/C2)
F = 643 N
Part B:
What is the proton's acceleration?
Express your answer to two significant figures and include the appropriate units.
SOLUTION:
We know that F=ma, so if we divide the force by the given mass... a = 3.8×1029 m/s2
The nucleus of a 125Xe atom (an isotope of the element xenon with mass 125 u) is 6.0 fm in diameter. It has 54 protons and charge q = +54e.
Part A:
What is the electric force on a proton 1.4 fm from the surface of the nucleus?
Hint: Treat the spherical nucleus as a point charge.
Express your answer to two significant figures and include the appropriate units.
SOLUTION:
We know the formula for the force between two charged particles is F = K |q1| |q2| r-2
We simply have to determine which values to plug into this equation and we will get the answer that we're looking for.
Begin by taking the radius of the nucleus, which is half the diameter, or 3 fm
the added distance of the proton is 1.4 fm, so the total distance of the proton from the center of the nucleus is 3 + 1.4 = 4.4 fm = 4.4⋅10-15 m
so r = 4.4⋅10-15 m
Next we need to determine the two charges. We're given that the nucleus has a charge of q = +54e.
Knowing that e=1.6⋅10-19 C , qnucleus = 8.64⋅10-18 C
q1 = 8.64⋅10-18 C
lastly, we know that the charge of ONE proton is just e, so
q2 = 1.6⋅10-19 C
Now we have all of our values to plug in and can solve for the force(remember K is the electrostatic constant and equal to 9.0⋅109 Nm2/C2)
∴F = (9.0⋅109 Nm2/C2)(8.64⋅10-18 C)(1.6⋅10-19 C)(4.4⋅10-15 m)-2
→F = 642.645 NPart B:
What is the proton's acceleration?
Express your answer to two significant figures and include the appropriate units.
SOLUTION:
We know that F=ma, so if we divide the force by the given mass... a = 3.8×1029 m/s2
Tuesday, December 9, 2014
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)
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^)
E = K ⋅ q/r2 ⋅ (r^)
Useful:
Electric constant
ε0 = 8.854×10-12 F/m
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:
FAonB= K⋅|qA⋅qB| / [a]2 = 1/[4πϵ0]⋅|(-1.0 nC)⋅(-2.0 nC)| / [2.0 cm]2 = 4.494×10-5N
FConB= K⋅|qC⋅qB| / 2= 1/[4πϵ0]⋅|(2.0 nC)⋅(-2.0 nC)| / [1.3 cm]2 = 2.13×10-4N
FNETonB = FAonB + FConB = 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)
FAonB is a negative charge pushing the negative charge down
FConB is a positive charge pulling the negative charge down
SO,
the force is directed down
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:
FAonB= K⋅|qA⋅qB| / [a]2 = 1/[4πϵ0]⋅|(-1.0 nC)⋅(-2.0 nC)| / [2.0 cm]2 = 4.494×10-5N
FConB= K⋅|qC⋅qB| / 2= 1/[4πϵ0]⋅|(2.0 nC)⋅(-2.0 nC)| / [1.3 cm]2 = 2.13×10-4N
FNETonB = FAonB + FConB = 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)
FAonB is a negative charge pushing the negative charge down
FConB 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
F1on3= 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
F2on3= 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)
F1on3 is a negative charge pulling the positive charge in the negative x-direction
F2on3 is a positive charge pushing the positive charge in the negative x-direction
SO, FNETon3 = F1on3(negative) + F2on3(negative) =-2.6×10-5 N + -1.388×10-5 N = -3.99×10-5 N
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
F1on3= 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
F2on3= 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)
F1on3 is a negative charge pulling the positive charge in the negative x-direction
F2on3 is a positive charge pushing the positive charge in the negative x-direction
SO, FNETon3 = F1on3(negative) + F2on3(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
EA = 40 N/C
rB = rA + 0.25 m = 0.5 m
EB = K ⋅ q/rB2
→[-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)]
EB = -10 N/C
what this means by -10 N/C is that it's pointing south, just as EA is
so ENET = -40 N/C + -10 N/C = -50 N/C
aka 50 N/C pointing south
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
EA = 40 N/C
rB = rA + 0.25 m = 0.5 m
EB = K ⋅ q/rB2
→[-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)]
EB = -10 N/C
what this means by -10 N/C is that it's pointing south, just as EA is
so ENET = -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)
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)
25: PSS 22.1 Electric Forces and Coulomb’s Law
INTRO:
Two charged particles, with charges q1=q and q2=4q, are located at a distance d=2.00cm apart on the x axis. A third charged particle, with charge q3=q, is placed on the x axis such that the magnitude of the force that charge 1 exerts on charge 3 is equal to the force that charge 2 exerts on charge 3.
Find the position of charge 3 when q = 2.00nC .
The mathematical representation is based on Coulomb’s law:
F1on2=F2on1=K⋅|q1|⋅|q2| / r2.
Part A: Which of the following sketches represents a possible configuration for this problem?
Enter the letter(s) indicating the correct graph(s) in alphabetical order. For example, if you think that A and B are correct, enter AB.
SOLUTION:
begin by looking and analyzing the given problem
for the forces to be equivalent, q3 must be significantly closer to q1 than to q2, because q2 is much stronger
However, q3 could technically be either between the charges or outside of them, as long as it obeys the rules we just stated.
sketch A could work because it follows our definition
sketch B can't work because q3 is much closer to q2 than to q1 so F2on3 would be much greater than F1on3
sketch C could work because it follows our definition
and sketch D couldn't work for the same reason sketch B can't.
So, our final answer is AC
Part B:
Suppose charge 3 is placed in between the other two charges, as shown below. Use this diagram to show the force vectors representing all the forces acting on charge 3.
Draw each vector with its tail at charge 3. The orientation of your vectors will be graded. The length of your vectors will not be graded.
SOLUTION:
this one is relatively difficult to type out, but I will try to.
both vectors point in toward charge 3
(q1)⇒ (q3) ⇐(q2)
NOTES:
When charge 3 is placed in between the other two charges, the electric forces exerted on it have opposite directions. In this configuration, when charge 3 is at a point that satisfies the conditions given in the problem, the net force on it will be zero because F2on3=F1on3. If, instead, charge 3 is located anywhere to the left of charge 1, both the electric forces exerted on it point to the left. Thus, the net force on charge 3 will also be directed to the left. You may want to add this information to your pictorial representation.
You are trying to find the position on the x axis of charge 3. Assuming charge 1 is located at the origin of the x axis and the positive x axis points to the right, label the position of charge 3 x3. Because of the two possible configurations, your calculations should yield two values for x3: a positive value x3,r when charge 3 is to the right of charge 1 and between the two charges, and a negative value x3,l when charge 3 is located to the left of charge 1 (that is, to the left of the origin).
Your final sketch should look like this:
Part C:
Assuming charge 1 is located at the origin of the x axis and the positive x axis points to the right, find the two possible values x3,r and x3,ℓ for the position of charge 3.
Express your answers in centimeters to three significant figures, separated by a comma.
SOLUTION:
givens:
q1 = q
q2 = 4q
q3 = q
d = 2.00 cm
xq1 = 0 cm
xq2 = 2.00 cm
q = 2.00 nC
next lets determine the forces
F1on3= K⋅|q1|⋅|q3| / r2 = K⋅|q|⋅|q| / (d1-3)2 = K⋅q2 / (d1-3)2
F2on3= K⋅|q2|⋅|q3| / r2 = K⋅|4q|⋅|q| / (d3-2)2 = K⋅4q2 / (d3-2)2
since the forces are supposed to equal each other in each case, we're able to solve for the variables in both cases
F1on3= F2on3
K⋅q2 / x2=K⋅4q2 / (2.00 cm - x)2
1/x2 = 4/ [4.00 cm2 - (4.00 cm⋅x) + x2]
[4.00 cm2 - (4.00 cm⋅x) + x2]/x2 = 4
4.00 cm2/x2 - (4.00 cm⋅x)/x2 +x2/x2 - 4 = 0
4.00 cm2/x2 - (4.00 cm)/x + 1 - 4 = 0
4.00 cm2/x2 - (4.00 cm)/x -3 = 0
3=4.00 cm2/x2 - (4.00 cm)/x
3x2=4.00 cm -4.00 cm⋅x
3x2 + 4.00 cm⋅x - 4.00 cm = 0
plugging that into the quadratic equation we get
x = -2 & x = 2/3
x3,r, x3,ℓ = 0.667,-2.00 cm, cm
Part D:
The key equation for this problem is F1on3=F2on3. In the previous part, you solved it by applying Coulomb's law and obtaining a quadratic equation with two real-valued solutions for x3. If, instead of writing an equation for x3, you apply Coulomb's law and write a simpler equation in terms of the distances of charge 3 from charges 1 and 2, respectively, r13 and r23, which of the following relations would you get?
a) r23=2r13
b) r23=½r13
c) r23=4r13
d) r23=¼r13
SOLUTION:
If we just backtrack to some of the math we did in the last part, we find
K⋅q2 / x2=K⋅4q2 / (2.00 cm - x)2
x2 is r13 and (2.00 cm - x)2 is r23
so, 1 /r132=4/ r232
→r232=4r132
→r23=2r13
or option a)
Two charged particles, with charges q1=q and q2=4q, are located at a distance d=2.00cm apart on the x axis. A third charged particle, with charge q3=q, is placed on the x axis such that the magnitude of the force that charge 1 exerts on charge 3 is equal to the force that charge 2 exerts on charge 3.
Find the position of charge 3 when q = 2.00nC .
The mathematical representation is based on Coulomb’s law:
F1on2=F2on1=K⋅|q1|⋅|q2| / r2.
Part A: Which of the following sketches represents a possible configuration for this problem?
SOLUTION:
begin by looking and analyzing the given problem
for the forces to be equivalent, q3 must be significantly closer to q1 than to q2, because q2 is much stronger
However, q3 could technically be either between the charges or outside of them, as long as it obeys the rules we just stated.
sketch A could work because it follows our definition
sketch B can't work because q3 is much closer to q2 than to q1 so F2on3 would be much greater than F1on3
sketch C could work because it follows our definition
and sketch D couldn't work for the same reason sketch B can't.
So, our final answer is AC
Part B:
Suppose charge 3 is placed in between the other two charges, as shown below. Use this diagram to show the force vectors representing all the forces acting on charge 3.
Draw each vector with its tail at charge 3. The orientation of your vectors will be graded. The length of your vectors will not be graded.
SOLUTION:
this one is relatively difficult to type out, but I will try to.
both vectors point in toward charge 3
(q1)⇒ (q3) ⇐(q2)
NOTES:
When charge 3 is placed in between the other two charges, the electric forces exerted on it have opposite directions. In this configuration, when charge 3 is at a point that satisfies the conditions given in the problem, the net force on it will be zero because F2on3=F1on3. If, instead, charge 3 is located anywhere to the left of charge 1, both the electric forces exerted on it point to the left. Thus, the net force on charge 3 will also be directed to the left. You may want to add this information to your pictorial representation.
You are trying to find the position on the x axis of charge 3. Assuming charge 1 is located at the origin of the x axis and the positive x axis points to the right, label the position of charge 3 x3. Because of the two possible configurations, your calculations should yield two values for x3: a positive value x3,r when charge 3 is to the right of charge 1 and between the two charges, and a negative value x3,l when charge 3 is located to the left of charge 1 (that is, to the left of the origin).
Your final sketch should look like this:
Part C:
Assuming charge 1 is located at the origin of the x axis and the positive x axis points to the right, find the two possible values x3,r and x3,ℓ for the position of charge 3.
Express your answers in centimeters to three significant figures, separated by a comma.
SOLUTION:
givens:
q1 = q
q2 = 4q
q3 = q
d = 2.00 cm
xq1 = 0 cm
xq2 = 2.00 cm
q = 2.00 nC
next lets determine the forces
F1on3= K⋅|q1|⋅|q3| / r2 = K⋅|q|⋅|q| / (d1-3)2 = K⋅q2 / (d1-3)2
F2on3= K⋅|q2|⋅|q3| / r2 = K⋅|4q|⋅|q| / (d3-2)2 = K⋅4q2 / (d3-2)2
since the forces are supposed to equal each other in each case, we're able to solve for the variables in both cases
F1on3= F2on3
K⋅q2 / x2=K⋅4q2 / (2.00 cm - x)2
1/x2 = 4/ [4.00 cm2 - (4.00 cm⋅x) + x2]
[4.00 cm2 - (4.00 cm⋅x) + x2]/x2 = 4
4.00 cm2/x2 - (4.00 cm⋅x)/x2 +x2/x2 - 4 = 0
4.00 cm2/x2 - (4.00 cm)/x + 1 - 4 = 0
4.00 cm2/x2 - (4.00 cm)/x -3 = 0
3=4.00 cm2/x2 - (4.00 cm)/x
3x2=4.00 cm -4.00 cm⋅x
3x2 + 4.00 cm⋅x - 4.00 cm = 0
plugging that into the quadratic equation we get
x = -2 & x = 2/3
x3,r, x3,ℓ = 0.667,-2.00 cm, cm
Part D:
The key equation for this problem is F1on3=F2on3. In the previous part, you solved it by applying Coulomb's law and obtaining a quadratic equation with two real-valued solutions for x3. If, instead of writing an equation for x3, you apply Coulomb's law and write a simpler equation in terms of the distances of charge 3 from charges 1 and 2, respectively, r13 and r23, which of the following relations would you get?
a) r23=2r13
b) r23=½r13
c) r23=4r13
d) r23=¼r13
SOLUTION:
If we just backtrack to some of the math we did in the last part, we find
K⋅q2 / x2=K⋅4q2 / (2.00 cm - x)2
x2 is r13 and (2.00 cm - x)2 is r23
so, 1 /r132=4/ r232
→r232=4r132
→r23=2r13
or option a)
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