Help with E-field integral please.

  • Thread starter Thread starter Fjolvar
  • Start date Start date
  • Tags Tags
    E-field Integral
Click For Summary

Homework Help Overview

The discussion revolves around solving an integral related to the electric field of a line charge in an electrodynamics context, specifically referencing Griffiths' textbook. The original poster expresses difficulty with the integral and seeks clarification on the integration techniques involved.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the integral of the form 1/[(z^2+x^2)]^(3/2) and various methods for solving it, including trigonometric substitution. Questions arise about the rules and identities used in integration, as well as the steps taken to arrive at the solution.

Discussion Status

Some participants have provided guidance on trigonometric substitution and shared their attempts to solve the integral. There is ongoing exploration of the integration process, with some participants expressing uncertainty about specific steps and seeking further clarification.

Contextual Notes

Participants mention a need to review calculus concepts, particularly trigonometric identities and substitution methods, indicating a potential gap in foundational knowledge that is relevant to the problem at hand.

Fjolvar
Messages
156
Reaction score
0
Hello, I need help solving an integral in the line charge equation for my electrodynamics class. I'm using Griffiths third edition and I understand where the components of the problem come from, but I can't seem to solve the integral, although I'm sure it's very basic. It's been awhile since I took calc 3 and I can't seem to find any similar examples online. I attached the solution to the problem and put a red box around the two integrals (Ez and Ex) that I need help understanding. Just the basic rules that were used to integrate and find the solution would be much appreciated. Thank you!
 

Attachments

  • Griffiths problem.jpg
    Griffiths problem.jpg
    25.1 KB · Views: 507
Physics news on Phys.org
Here it is written out..

Integral of 1/[(z^2+x^2)]^(3/2) dx. How do I integrate this?
 
[tex]\int \frac{dx}{(x^2+z^2)^{3/2}} = \frac{x}{z^2 \sqrt{x^2+z^2}} + C[/tex]
 
I know the solution, but I wanted to know how to get it. What rules did you use?
 
trig substitution x= ztan( theta).
 
Hmm, is there a list of these identities I can refer to? I'm still not seeing it heh.
 
Can anyone please elaborate on the steps taken to solve this integral?
 
If you do not know how to use trigonometric substitutions you can pretty much forget about getting an answer to your question. I suggest you review your calc text or use your best friend, google.
 
Alright so I tried to relearn trig sub. and attempted the problem but I'm going wrong somewhere. I scanned my work and attached it. If someone can quickly review and perhaps point out my mistake it would really mean a lot. Thanks.
 

Attachments

  • #10
If [tex]x = z \tan \theta[/tex] then [tex]dx = z \sec^2 \theta \mathbf{d \theta}[/tex]
 
  • #11
Okay I finally figured out how to get the answer. Sorry to keep this thread going! At least I feel caught up on trigonometric substitution now. Thanks ╔(σ_σ)╝ and Inferior89.
 
Last edited:
  • #12
Integrate and substitute back from [tex]\theta[/tex]
[tex]\sin ( \arctan (x/z) ) = \frac{x}{\sqrt{x^2 + z^2}}[/tex]
 
  • #13
I have another question about this problem. In the solution above they calculate each component of the vector field Ez and Ex by using cos theta and sin theta in place of the directional r hat vector. This may be a stupid question, but how do you determine whether to use cosine or sine in relation to a vector component.. such as Ex, Ey, or Ez? I'm comparing to the general equation used to find an Electric field of a line charge, E = (1/4pi epsilon o) * (r hat * lambda/ r^2) where they replaced r hat by cos theta.. Thank you.
 
  • #14
Anyone?
 

Similar threads

Replies
7
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
Replies
2
Views
1K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 24 ·
Replies
24
Views
4K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 6 ·
Replies
6
Views
4K