Electric field of a wire segment

  • Thread starter Reshma
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  • #1
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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
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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
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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
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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
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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
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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
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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]
 

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