Work on a charged particle due to Ring

In summary, the problem involves finding the work required and maximum speed of moving a charged particle from far away to the center of a ring with a fixed diameter and uniform charge distribution. The potential energy is used to calculate the work, and conservation of energy is used to find the maximum speed. There is no need for integration to find the electric field or potential at a point in front of the ring, as the magnitude and direction of the electric field are the same at all points on the axis. Integration is only required for calculating the electric field at a point off the axis.
  • #1
dwstrait
4
0

Homework Statement


A ring of diameter 7.90 cm is fixed in place and carries a charge of 5.90 [tex]\mu[/tex]C uniformly spread over its circumference. How much work does it take to move a tiny 3.80 [tex]\mu[/tex]C charged ball of mass 1.90 g from very far away to the center of the ring? Also, What is the maximum speed it will reach?

Homework Equations



W = -[tex]\Delta[/tex]U
U(r) = (kqq0)/r
KE + PE = 0

The Attempt at a Solution



In order to get the work it takes to move the particle, I figured I could find the change in the Potential Energy. The potential energy when the particle is very far away is zero. Is there a better way to approach the problem than using PE?

In order to find velocityf, I can just use conservation of energy and solve for velocity. So, KEf = sqrt{-2PEf/m}
 
Last edited:
Physics news on Phys.org
  • #2
This looks like quite a difficult problem. Unless you have a potential formula designed for rings, you will have to use integration to sum the contributions to potential from all the point charges dq on the ring.
 
  • #3
Oh, no need to integrate because the point you are interested in is equally distant from all of the charge! Sorry I misled you - not a difficult problem at all!

I don't understand the second part, maximum speed. Surely that just depends on how it is pushed?
 
  • #4
Thanks for suggesting that I didn't need to integrate. I was stuck trying to work out an integral, but once I saw I didn't need one, the problem was very simple.

For the maximum speed, I just solved KE + PE = 0 for velocity, and that was correct.
 
  • #5
I have a similar problem, and I get the right answer without integrating, which is great! But that confuses me. There is a whole section(Halliday and Resnik 8th Extended Edition, page 587- The Electric Field Due To A Line Of Charge) that details on how to integrate to get the electric field at a point P, at a distance, z, from the ring. Why do I need to integrate to get the electric field at a point positioned somewhere in front of the ring, yet I don't have to if I want the potential on a particle at the same point.

(by the way I hope I'm not breaking a rule by posting this here, but it seemed to be pretty pointless starting a new thread and explaining the problem again.)
 
  • #6
Maybe I'm missing something, but if the title of the section is "the electric field due to a line of charge", shouldn't the section talk about lines of charge instead of rings? You don't need to integrate to find the electric field along the axis of a ring. You also don't need integration for finding the field due to a line of charge.
 
  • #7
K29 said:
I have a similar problem, and I get the right answer without integrating, which is great! But that confuses me. There is a whole section(Halliday and Resnik 8th Extended Edition, page 587- The Electric Field Due To A Line Of Charge) that details on how to integrate to get the electric field at a point P, at a distance, z, from the ring. Why do I need to integrate to get the electric field at a point positioned somewhere in front of the ring, yet I don't have to if I want the potential on a particle at the same point.
The different elements of ring produce the electric field at a point having the same magnitude but different direction. Electric field is a vector quantity which you cannot add directly. Whereas the electric potential is a scalar quantity which you can add directly. Therefore we require integration to find the electric field.
 
  • #8
ah thanks. Fundamentals, fundamentals, fundamentals! :)
 

1. What is work on a charged particle due to a ring?

The work on a charged particle due to a ring is the amount of energy required to move the particle from one point to another in the presence of a ring-shaped electric field.

2. How is the work on a charged particle calculated?

The work on a charged particle can be calculated using the equation W = qΔV, where W is the work, q is the charge of the particle, and ΔV is the change in electric potential between the two points.

3. What factors affect the work on a charged particle due to a ring?

The work on a charged particle due to a ring is affected by the charge of the particle, the strength of the electric field, and the distance between the particle and the ring.

4. Can the work on a charged particle be positive or negative?

Yes, the work on a charged particle can be positive or negative depending on the direction of the electric field and the direction of the particle's movement. A positive work means that the electric field does positive work on the particle, moving it in the same direction as the field, while a negative work means that the electric field does negative work on the particle, moving it in the opposite direction of the field.

5. What are some real-world applications of work on a charged particle due to a ring?

The concept of work on a charged particle due to a ring is used in various applications such as particle accelerators, mass spectrometers, and ion traps. It is also important in understanding the behavior of charged particles in electromagnetic fields, which has implications in fields such as plasma physics, astrophysics, and atmospheric sciences.

Similar threads

  • Introductory Physics Homework Help
Replies
3
Views
2K
  • Introductory Physics Homework Help
Replies
9
Views
2K
  • Introductory Physics Homework Help
Replies
16
Views
2K
  • Introductory Physics Homework Help
Replies
13
Views
1K
  • Introductory Physics Homework Help
Replies
2
Views
1K
  • Introductory Physics Homework Help
Replies
4
Views
1K
  • Introductory Physics Homework Help
Replies
3
Views
1K
  • Introductory Physics Homework Help
Replies
3
Views
1K
  • Introductory Physics Homework Help
Replies
7
Views
975
  • Introductory Physics Homework Help
Replies
1
Views
673
Back
Top