Electric field of a wire segment

  • Thread starter Reshma
  • Start date
  • #1
Reshma
749
4
This is a solved example in Griffith's book.
Question: Find the electirc field a distance 'z' above the midpoint of a straight line segment of length 2L, which carries uniform charge [itex]\lambda[/itex]

Solution: Let me work out the steps and show you where my problem is(I have avoided showing some of the steps).
The coordinate axis has been set up taking the midpoint of the wire as the origin. Magnitude of the electric field is given by:
[tex] E = \frac{1}{4\pi \epsilon_0} \int_{0}^{L} \frac{2\lambda z}{(z^2 + x^2)^{3/2}} dx[/tex]

HERE, Griffith has not elaborated the evauation of the integral under the bracket and directly given the solution. I have tried my level best to solve it but couldn't suceed. So can someone show me the intermediate steps involved in evaluating the integral:
[tex]\int_{0}^{L}\frac{1}{(z^2 + x^2)^{3/2}} dx[/tex] ([itex]\lambda[/itex] & z are constants)

For your convenience, the final solution is:
[tex]E = \frac{1}{4\pi \epsilon_0} [\frac{2\lambda L}{z\sqrt{z^2 + L^2}}][/tex]
 

Answers and Replies

  • #2
Kelvin
52
0
to deal with the integral [tex]\int \frac{1}{(z^2+x^2)^{3/2}} dx[/tex]
try putting [tex]x = z \tan\theta[/tex]

and using the identity [tex]1+\tan^2\theta = \sec^2\theta[/tex]
 
  • #3
Reshma
749
4
Kelvin said:
to deal with the integral [tex]\int \frac{1}{(z^2+x^2)^{3/2}} dx[/tex]
try putting [tex]x = z \tan\theta[/tex]

and using the identity [tex]1+\tan^2\theta = \sec^2\theta[/tex]

Thank you for the guidance. Here is my evalution of the integral:
[tex]\int_{0}^{L} \frac{z^2\sec^2 \theta d\theta}{({z^2\tan^2 \theta +z^2})^{3/2}}[/tex]
[tex]= \frac{1}{z^2} \int_{0}^{L} \cos \theta d\theta[/tex]
[tex]= \frac{1}{z^2} [\sin \theta] \mid_{0}^{L}[/tex]
[tex]= \frac{1}{z^2} \frac{x}{\sqrt{z^2 + x^2}} \mid_{0}^{L}[/tex]
[tex]=\frac{1}{z^2} \frac{L}{\sqrt{z^2 + L^2}}[/tex]
 
  • #4
Reshma
749
4
Sticking to the original question, electric field is:

[tex]E = \frac{1}{4\pi \epsilon_0} [\frac{2\lambda L}{z\sqrt{z^2 + L^2}}][/tex]
How will the formula be modified if [itex]L\rightarrow \infty[/itex]
 
  • #5
Doc Al
Mentor
45,447
1,907
Consider that [tex]\sqrt{z^2 + L^2} = L \sqrt{1 + (z/L)^2}[/tex]

Now can you see what happens when [itex]L\rightarrow \infty[/itex] ?
 
  • #6
Reshma
749
4
Doc Al said:
Consider that [tex]\sqrt{z^2 + L^2} = L \sqrt{1 + (z/L)^2}[/tex]

Now can you see what happens when [itex]L\rightarrow \infty[/itex] ?
Wow! Thanks Doc Al :smile:.

[tex]E = \frac{1}{4\pi \epsilon_0} [\frac{2\lambda L}{z L\sqrt{1 + (z/L)^2}}][/tex]

So,[itex](z/L)^2\rightarrow 0[/itex]

Hence,
[tex]E = \frac{2\lambda}{4\pi \epsilon_0 z}[/tex]
 
  • #7
xts
882
0
Please note, that the solution in a limit of L→∞ is nonphysical. The total charge of such system is infinite. That leads to another singularity: the electrostatic potential at any point of the space is also infinite.
[itex]\varphi(z)=\int_{z}^{\infty}E(\zeta)d\zeta=\infty[/itex]
 

Suggested for: Electric field of a wire segment

Replies
10
Views
110
  • Last Post
Replies
6
Views
394
Replies
1
Views
364
  • Last Post
Replies
7
Views
113
  • Last Post
Replies
4
Views
258
Replies
5
Views
395
Replies
21
Views
409
Replies
14
Views
458
Replies
21
Views
147
Top