26: Charged Ring

INTRO:
Consider a uniformly charged ring in the xy plane, centered at the origin. The ring has radius a and positive charge q distributed evenly along its circumference. (Figure 1)

Figure 1

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PART A:
What is the direction of the electric field at any point on the z axis?

• parallel to the x axis
• parallel to the y axis
• parallel to the z axis
• in a circle parallel to the xy plane

SOLUTION:
Parallel to the z-axis

NOTES:
Approach 1
In what direction is the field due to a point on the ring? Add to this the field from a point on the opposite side of the ring. In what direction is the net field? What if you did this for every pair of points on opposite sides of the ring?

Approach 2
Consider a general electric field at a point on the z axis, i.e., one that has a z component as well as a component in the xy plane. Now imagine that you make a copy of the ring and rotate this copy about its axis. As a result of the rotation, the component of the electric field in the xy plane will rotate also. Now you ask a friend to look at both rings. Your friend wouldn't be able to tell them apart, because the ring that is rotated looks just like the one that isn't. However, they have the component of the electric field in the xy plane pointing in different directions! This apparent contradiction can be resolved if this component of the field has a particular value. What is this value?
Does a similar argument hold for the z component of the field?
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PART B:
What is the magnitude of the electric field along the positive z axis?
Use k in your answer, where k=1/(4πϵ₀).

SOLUTION:
Use Coulomb's law, F=k⋅q₁⋅q₂/r², to find the electric field (the Coulomb force per unit charge) due to a point charge. 
Given the force, the electric field at q₂ due to q₁ is E=F/q²=k⋅q₁/r².

Next, use Coulomb's law to find the contribution dE to the electric field at the point (0,0,z) from a piece of charge dq on the ring at a distance r away. Then, you can integrate over the ring to find the value of E. Consider an infinitesimal piece of the ring with charge dq. Use Coulomb's law to write the magnitude of the infinitesimal dE at a point on the positive z axis due to the charge dq shown in the figure.

∴ dE = kdq/(z² + a²)

By symmetry, the net field must point along the z axis, away from the ring, because the horizontal component of each contribution of magnitude dE is exactly canceled by the horizontal component of a similar contribution of magnitude dE from the other side of the ring. Therefore, all we care about is the z component of each such contribution. 
The component dE_z of the electric field caused by the charge on an infinitesimally small portion of the ring dq in the z direction is... 
dE_z = dE_z/(z² + a²)^½

Next, integrate around the ring. 
By combining the previous two results, you will have an expression for dE_z, 
the vertical component of the field due to the infinitesimal charge dq. 

The total field is
E =E_zk^=k^∮_ringdE_z.


If you are not comfortable integrating dq over the ring, change to a spatial variable. Since the total charge q is distributed evenly about the ring, convince yourself that

∮_ringdq=∫₀^2πq/(2π) dθ.

->    E(z) = kzq/(z² + a²)^3/2

NOTE: 
Notice that this expression is valid for both positive and negative charges as well as for points located on the positive and negative z axis. If the charge is positive, the electric field should point outward. For points on the positive z axis, the field points in the positive z direction, which is outward from the origin. For points on the negative z axis, the field points in the negative z direction, which is also outward from the origin. If the charge is negative, the electric field should point toward the origin. For points on the positive z axis, the negative sign from the charge causes the electric field to point in the negative z direction, which points toward the origin. For points on the negative z axis, the negative sign from the z coordinate and the negative sign from the charge cancel, and the field points in the positive z direction, which also points toward the origin. Therefore, even though we obtained the above result for postive q and z, the algebraic expression is valid for any signs of the parameters. As a check, it is good to see that if |z| is much greater than a the magnitude of E(z) is approximately kq/(z²), independent of the size of the ring: The field due to the ring is almost the same as that due to a point charge q at the origin.

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PART C:

Imagine a small metal ball of mass m and negative charge −q₀. The ball is released from rest at the point (0,0,d) and constrained to move along the z axis, with no damping. If 0<d≪a, what will be the ball's subsequent trajectory?

• repelled from the origin
• attracted toward the origin and coming to rest
• oscillating along the z axis between z=d and z=−d
• circling around the z axis at z=d

SOLUTION:

oscillating along the z axis between z=d and z=−d

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PART D:

The ball will oscillate along the z axis between z=d and z=−d in simple harmonic motion. What will be the angular frequency ω of these oscillations? Use the approximation d≪a to simplify your calculation; that is, assume that d²+a²≈a².
Express your answer in terms of given charges, dimensions, and constants.

SOLUTION:

Recall the nature of simple harmonic motion of an object attached to a spring. Newton's second law for the system states that
F_x=m⋅d²x/dt²=−k′x, leading to oscillation at a frequency of ω=√k′/m

(here, the prime on the symbol representing the spring constant is to distinguish it from k=1/(4πϵ₀). The solution to this differential equation is a sinusoidal function of time with angular frequency ω. Write an analogous equation for the ball near the charged ring in order to find the ω term.

Next find the force on the charge
What is F_z, the z component of the force on the ball on the ball at the point (0,0,d)? Use the approximation d²+a²≈a².

F_z = (-k⋅q⋅q₀⋅d)/a³

^-->     ω = [(k⋅q⋅q₀)/(a³⋅m)]^½

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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

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 , q_nucleus = 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 N​

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

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: 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:
F_1on2=F_2on1=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 F_2on3 would be much greater than F_1on3
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 F_2on3=F_1on3. 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 x_3,r when charge 3 is to the right of charge 1 and between the two charges, and a negative value x_3,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 x_3,r and x_3,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
x_q1 = 0 cm
x_q2 = 2.00 cm
q = 2.00 nC

next lets determine the forces
F_1on3= K⋅|q₁|⋅|q₃| / r² = K⋅|q|⋅|q| / (d_1-3)² = K⋅q² / (d_1-3)²
F_2on3= K⋅|q₂|⋅|q₃| / r² = K⋅|4q|⋅|q| / (d_3-2)² = K⋅4q² / (d_3-2)²

since the forces are supposed to equal each other in each case, we're able to solve for the variables in both cases

F_1on3= F_2on3
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 
x_3,r, x_3,ℓ = 0.667,-2.00 cm, cm

Part D:
The key equation for this problem is F_1on3=F_2on3. 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)