Evaluating a simple electrostatic field integral

In summary, the conversation discusses finding the e-field at point P and the correct integral solution for the y-component, Ey. The question asks for a more rigorous explanation, and L'Hôpital's Rule is suggested but does not lead to a definitive answer. The conversation concludes by discussing factoring and correctly factoring the denominator to find the limit as x approaches infinity.
  • #1
Taulant Sholla
96
5

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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  • #2
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
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
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.
 
  • #5
Ach, yes. Sorry - and thank you!
 
  • #6
Taulant Sholla said:
Ach, yes. Sorry - and thank you!
Kein Problem!
 

1. What is a simple electrostatic field integral?

A simple electrostatic field integral is a mathematical representation of the electric field surrounding a charged object. It calculates the amount of work required to move a test charge from one point to another in the presence of an electric field.

2. How is a simple electrostatic field integral evaluated?

A simple electrostatic field integral is evaluated by using the formula ∫E·dl = q/ε0, where E is the electric field, dl is the infinitesimal displacement along the path of integration, q is the charge of the object, and ε0 is the permittivity of free space.

3. What factors affect the value of a simple electrostatic field integral?

The value of a simple electrostatic field integral can be affected by the magnitude and distribution of the electric charge, the distance between the charged object and the test charge, and the orientation of the electric field with respect to the path of integration.

4. What are some real-life applications of evaluating simple electrostatic field integrals?

Simple electrostatic field integrals are used in a variety of real-life applications, such as in the design of electronic devices, the study of lightning and other atmospheric phenomena, and the development of medical imaging techniques.

5. Are there any limitations to using simple electrostatic field integrals?

Yes, there are some limitations to using simple electrostatic field integrals. They assume a steady-state electric field, which may not be the case in dynamic systems. They also do not take into account the effects of magnetic fields or the presence of other charges in the surrounding space.

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