Ring of Charge, Electric Potential

Click For Summary
SUMMARY

The discussion focuses on calculating the initial speed required for a proton to cross the plane of a uniformly charged thin ring with a charge density of 5.0 nC/cm and a radius of 1.0 cm. The proton starts 2.0 cm from the center of the ring and moves towards it. Key equations include ΔU = ΔV*q, ΔKE + ΔU = 0, and Vcenter = KQ/R. Participants emphasize the importance of calculating the total charge using the ring's circumference and applying the Pythagorean theorem to determine distances accurately.

PREREQUISITES
  • Understanding of electric potential and potential energy
  • Familiarity with the concepts of charge density and total charge
  • Knowledge of the Pythagorean theorem for distance calculations
  • Basic principles of kinetic and potential energy in physics
NEXT STEPS
  • Learn how to calculate electric potential using the formula V = KQ/R
  • Study the relationship between potential energy and kinetic energy in electrostatics
  • Explore the concept of charge density and its implications in electric fields
  • Investigate the application of the Pythagorean theorem in three-dimensional charge distributions
USEFUL FOR

Students studying electromagnetism, physics educators, and anyone interested in electrostatic potential and motion of charged particles in electric fields.

fornax
Messages
20
Reaction score
0

Homework Statement


A proton is moving along the main axis of a uniformly charged thin ring. The charge density on the ring is 5.0nC/cm and the ring radius is 1.0cm. Initially the proton is 2.0cm (along the axis) from the center of the ring with the velocity towards the center of the ring. What initial speed should the proton have such it will cross the plane of the ring?


Homework Equations


ΔU = ΔV*q
ΔKE + ΔU = 0
Vcenter = KQ/R
W = -ΔU


The Attempt at a Solution



50nC/cm = 500nC/m

Vcenter = KQ/R
V = ((9x10^9)*(500x10^-9))/.01^2
V = 4.5x10^7

ok, looking at my paper, I then went on to multiply V*d, and I'm not sure why anymore...
I can't seem to find the correct equation for the V on the axis in my notes. I could use V*q for ΔU then user ΔU to ger ΔKE(1/2mv^2) but that gets me velocity if it accelerated toward the ring... ugh.

If I had V at .02m, the I could get ΔV between the center and there, then get ΔU, assume ΔKE is zero at the center, then get the velcoity at .02m with ΔU = ΔKE. Am I reasoning this correctly? Also, mind sharing that formula? :)
 
Physics news on Phys.org
fornax said:

Homework Statement


A proton is moving along the main axis of a uniformly charged thin ring. The charge density on the ring is 5.0nC/cm and the ring radius is 1.0cm. Initially the proton is 2.0cm (along the axis) from the center of the ring with the velocity towards the center of the ring. What initial speed should the proton have such it will cross the plane of the ring?

Homework Equations


ΔU = ΔV*q
ΔKE + ΔU = 0
Vcenter = KQ/R
W = -ΔU

The Attempt at a Solution



5[STRIKE]0[/STRIKE]nC/cm = 500nC/m

Vcenter = KQ/R
That looks about right to me so far, excluding the typo in bold and crossed out. (5.0 nC/cm, not 50 nC/cm) :approve:
V = ((9x10^9)*(500x10^-9))/.01^2
Okay, above in red is 'K'. So far so good. :smile:
V = ((9x10^9)*(500x10^-9))/.01^2
Above in green is supposed to be the total charge 'Q'. But 500 nC/m is the charge per unit length. You need to multiply that by the length of the ring to get the total charge [Edit: By 'length' here, I am referring to the ring's 'circumference'].
V = ((9x10^9)*(500x10^-9)/.01^2
Why are you squaring the distance?

(Your goal is to find the electric potential, not the electric force.)
ok, looking at my paper, I then went on to multiply V*d, and I'm not sure why anymore...
Nor am I. :rolleyes:
I can't seem to find the correct equation for the V on the axis in my notes. I could use V*q for ΔU then user ΔU to ger ΔKE(1/2mv^2) but that gets me velocity if it accelerated toward the ring... ugh.
That would give you the potential energy with respect to infinity.
If I had V at .02m, the I could get ΔV between the center and there, then get ΔU, assume ΔKE is zero at the center, then get the velcoity at .02m with ΔU = ΔKE. Am I reasoning this correctly?
That's exactly the correct reasoning (if I'm understanding you correctly). :approve:

(Just make sure that when you calculate the electric potential at a distance of 0.02 m [from the center of the ring], realize that the distance to the charge involves the Pythagorean theorem. :wink:)
Also, mind sharing that formula? :)

The Pythagorean theorem, I'm sure you already know. You already have the equation for electrical potential (with respect to infinity) in your relevant equations.
 
Last edited:

Similar threads

Replies
11
Views
4K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 11 ·
Replies
11
Views
5K
  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
5
Views
1K
  • · Replies 15 ·
Replies
15
Views
2K
  • · Replies 9 ·
Replies
9
Views
8K