# Evaluating a simple electrostatic field integral

1. Feb 20, 2017

### Taulant Sholla

1. The problem statement, all variables and given/known data
I can find the e-field at point P.

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

which I can see, informally, evaluates to:

3. 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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2. Feb 20, 2017

### blue_leaf77

Try factoring out $x^2$ from the squareroot in the denominator, you should know how the resulting $R^2/x^2$ behaves when $x \to \infty$.

3. Feb 20, 2017

### Taulant Sholla

Hmmm (thanks!). Yes, R2/x2 disappears as x→∞, but that now leaves an un-cancelled x in the numerator that is going to infinity.

4. Feb 20, 2017

### rude man

You're not factoring the denominator correctly.
(x2 + R2)1/2 = {x2(1 + R2/x2)}1/2
= x(1 + R2/x2)1/2
etc.

5. Feb 20, 2017

### Taulant Sholla

Ach, yes. Sorry - and thank you!

6. Feb 21, 2017

### rude man

Kein Problem!