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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