Help with E-field integral please.

In summary: I don't remember which direction to use when calculating a vector component. You would have to consult a textbook or look it up online.
  • #1
Fjolvar
156
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: 462
Physics news on Phys.org
  • #2
Here it is written out..

Integral of 1/[(z^2+x^2)]^(3/2) dx. How do I integrate this?
 
  • #3
[tex] \int \frac{dx}{(x^2+z^2)^{3/2}} = \frac{x}{z^2 \sqrt{x^2+z^2}} + C [/tex]
 
  • #4
I know the solution, but I wanted to know how to get it. What rules did you use?
 
  • #5
trig substitution x= ztan( theta).
 
  • #6
Hmm, is there a list of these identities I can refer to? I'm still not seeing it heh.
 
  • #7
Can anyone please elaborate on the steps taken to solve this integral?
 
  • #8
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.
 
  • #9
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 alot. Thanks.
 

Attachments

  • Integral Problem.pdf
    639 KB · Views: 242
  • #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?
 

FAQ: Help with E-field integral please.

What is an E-field integral?

An E-field integral is a mathematical tool used to calculate the strength and direction of an electric field at a given point in space.

How do I calculate an E-field integral?

To calculate an E-field integral, you will need to determine the electric field at each point along a specified path and then integrate the values using the appropriate mathematical formula.

What are the units of an E-field integral?

The units of an E-field integral are typically Newton-meters per coulomb (N·m/C) or volts per meter (V/m).

What is the significance of an E-field integral?

An E-field integral is important in understanding the behavior of electric fields and their effects on charged particles. It is also used in various engineering and scientific applications, such as designing electrical circuits.

Are there any limitations to using an E-field integral?

Yes, there are limitations to using an E-field integral. It assumes a static electric field and does not account for changing fields or the effects of magnetic fields. It also requires an accurate representation of the electric field along the specified path.

Similar threads

Replies
5
Views
1K
Replies
6
Views
1K
Replies
24
Views
2K
Replies
3
Views
1K
Replies
4
Views
1K
Replies
6
Views
2K
Back
Top