Work on a charged particle due to Ring

AI Thread Summary
A fixed ring with a diameter of 7.90 cm carries a uniform charge of 5.90 μC, and the problem involves calculating the work needed to move a 3.80 μC charged ball to its center. The potential energy at a distance is zero, simplifying the calculation of work as the change in potential energy. The maximum speed of the charged ball can be determined using conservation of energy principles, without the need for integration due to the symmetry of the ring's charge distribution. The discussion highlights the difference between calculating electric fields, which requires integration due to vector nature, and electric potential, which can be summed directly as a scalar. Understanding these fundamentals clarifies the approach to solving the problem effectively.
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Homework Statement


A ring of diameter 7.90 cm is fixed in place and carries a charge of 5.90 \muC uniformly spread over its circumference. How much work does it take to move a tiny 3.80 \muC 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 = -\DeltaU
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}
 
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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.
 
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?
 
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.
 
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.)
 
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.
 
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.
 
ah thanks. Fundamentals, fundamentals, fundamentals! :)
 
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