Evaluating a simple electrostatic field integral

AI Thread Summary
The discussion revolves around evaluating the electric field at a specific point using an integral solution for the y-component. The original poster successfully finds the correct answer but questions whether their explanation of canceling terms at infinity is sufficient. They attempted to use L'Hôpital's Rule but found it unhelpful. Other participants suggest factoring the denominator correctly to clarify the behavior of the terms as x approaches infinity. The conversation emphasizes the importance of rigorous mathematical justification in the evaluation process.
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!
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