Evaluating a simple electrostatic field integral

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Homework Help Overview

The discussion revolves around evaluating an electrostatic field integral, specifically focusing on the y-component of the electric field at a point P. Participants are exploring the mathematical behavior of the integral as certain variables approach infinity.

Discussion Character

  • Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster questions whether a simple explanation regarding the cancellation of terms at infinity is sufficient or if a more rigorous approach is necessary. They mention attempting L'Hôpital's Rule without definitive results.

Discussion Status

Some participants have provided hints regarding factoring terms in the integral to analyze their behavior as variables approach infinity. There is an ongoing exploration of how different terms behave under these limits, with no explicit consensus reached yet.

Contextual Notes

Participants are discussing the implications of certain mathematical manipulations and the rigor required in their explanations, indicating a focus on the correctness of their approaches rather than arriving at a final answer.

Taulant Sholla
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Homework Statement


I can find the e-field at point P.
point p.JPG


Homework Equations


I get, easily enough, the correct integral solution (for the y-component, Ey - which is all I need to do):
p4.JPG

which I can see, informally, evaluates to:

p3.JPG

which is the correct answer.

The Attempt at a Solution


My question: is it, technically, sufficient for me to simply explain: "well, when evaluated, the infinity in the numerator cancels with sqrt(infinity^2) in the denominator? Or can I offer a more rigorous explanation?

I tried L'Hôpital's Rule, but it doesn't seem to lead anywhere as far as showing anything definitive.

Thank you - any hints are appreciated!
 

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Try factoring out ##x^2## from the squareroot in the denominator, you should know how the resulting ##R^2/x^2## behaves when ##x \to \infty##.
 
Hmmm (thanks!). Yes, R2/x2 disappears as x→∞, but that now leaves an un-cancelled x in the numerator that is going to infinity.
 
Taulant Sholla said:
Hmmm (thanks!). Yes, R2/x2 disappears as x→∞, but that now leaves an un-cancelled x in the numerator that is going to infinity.

You're not factoring the denominator correctly.
(x2 + R2)1/2 = {x2(1 + R2/x2)}1/2
= x(1 + R2/x2)1/2
etc.
 
Ach, yes. Sorry - and thank you!
 
Taulant Sholla said:
Ach, yes. Sorry - and thank you!
Kein Problem!
 

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