Electric field of a cylinder given the electric field of a ring

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Homework Help Overview

The discussion revolves around understanding the electric field generated by a cylinder, particularly in relation to the electric field of a ring of charge. Participants are exploring the mathematical setup and integration process involved in deriving the electric field expression.

Discussion Character

  • Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants question the introduction of the variable z’ for integration, seeking clarity on its necessity compared to using z. There is also confusion regarding the absence of the radius R in the final expression and its implications for dimensional correctness. Additionally, questions arise about the integration limits and the treatment of the electric field direction in relation to the angle theta.

Discussion Status

The discussion is active, with participants engaging in clarifying the mathematical reasoning behind the integration process. Some guidance has been offered regarding the roles of different variables, but multiple interpretations and questions remain unresolved.

Contextual Notes

Participants are navigating potential typos in the provided solutions and are considering the implications of integrating over different variables. There is an emphasis on understanding the differences between the contributions from rings in the context of a disk versus a cylinder.

notgoodatphysics
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Homework Statement
We’re given the equation for the electric field of a disk. From that the idea is to find the electric field of a cylinder.

I thought the best way would be to integrate the original equation over the surface area of a cylinder without the ends (2*pi*r*h). My attempt is similar to the solution except, the professor has introduced z’, and an R in the final solution has disappeared.

Why introduce z’ and have the dz’ above the origin instead of just using z like the original diagram?

And where did the R go in the numerator in final step of the solution?
Relevant Equations
The first pic is the question and my attempt, and the second pic is the solution.
 
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notgoodatphysics said:
Why introduce z’ and have the dz’ above the origin instead of just using z like the original diagram?
The symbol ##z## represents the position of the point on the ##z##-axis. It's fixed. You can't integrate over that variable. You need a different variable, namely ##z'##, which corresponds to the position of an infinitesimal ring. Then you integrate with respect to ##z'## to sum the contributions over the entire cylinder.

notgoodatphysics said:
And where did the R go in the numerator in final step of the solution?
I think this is just a typo, and the ##R## should still be there. The final answer in your professor's solution isn't dimensionally correct.
 
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vela said:
The symbol ##z## represents the position of the point on the ##z##-axis. It's fixed. You can't integrate over that variable. You need a different variable, namely ##z'##, which corresponds to the position of an infinitesimal ring. Then you integrate with respect to ##z'## to sum the contributions over the entire cylinder.I think this is just a typo, and the ##R## should still be there. The final answer in your professor's solution isn't dimensionally correct.
Thanks for the reply!

I’m not sure I’m totally understanding the need for z’ though. In the example of the electric field of due to a ring of charge with radius a, to find the electric charge due to a disk, the integral from 0 to R is calculated - isn’t this a similar case? Why isn’t it (a-R).

(From here: https://www.physics.udel.edu/~watson/phys208/exercises/kevan/efield1.html )

Also when we find the integral of (z-z’), why don’t we also take the direction of the electric field E(z) ##cos theta## because ##\cos \theta##is changing right?
 
notgoodatphysics said:
I’m not sure I’m totally understanding the need for z’ though. In the example of the electric field of due to a ring of charge with radius a, to find the electric charge due to a disk, the integral from 0 to R is calculated - isn’t this a similar case? Why isn’t it (a-R).
When you break up a disk into a collection of rings, what's different about each ring? Similarly, when you divide the cylinder up into a collection of rings, what's different about those rings?
 

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