Electric field of a wire segment

AI Thread Summary
The discussion focuses on finding the electric field a distance 'z' above the midpoint of a uniformly charged wire segment of length 2L. The electric field is derived using an integral that requires evaluating the expression involving the distance and charge density. A specific substitution, x = z tan(θ), is suggested to simplify the integral. As L approaches infinity, the electric field formula modifies to E = (2λ)/(4πε₀z), but this scenario is deemed nonphysical due to the infinite total charge leading to an infinite electrostatic potential. The conversation highlights the importance of careful evaluation of integrals in electrostatics.
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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 \lambda

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:
E = \frac{1}{4\pi \epsilon_0} \int_{0}^{L} \frac{2\lambda z}{(z^2 + x^2)^{3/2}} dx

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:
\int_{0}^{L}\frac{1}{(z^2 + x^2)^{3/2}} dx (\lambda & z are constants)

For your convenience, the final solution is:
E = \frac{1}{4\pi \epsilon_0} [\frac{2\lambda L}{z\sqrt{z^2 + L^2}}]
 
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to deal with the integral \int \frac{1}{(z^2+x^2)^{3/2}} dx
try putting x = z \tan\theta

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

and using the identity 1+\tan^2\theta = \sec^2\theta

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

E = \frac{1}{4\pi \epsilon_0} [\frac{2\lambda L}{z\sqrt{z^2 + L^2}}]
How will the formula be modified if L\rightarrow \infty
 
Consider that \sqrt{z^2 + L^2} = L \sqrt{1 + (z/L)^2}

Now can you see what happens when L\rightarrow \infty ?
 
Doc Al said:
Consider that \sqrt{z^2 + L^2} = L \sqrt{1 + (z/L)^2}

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

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

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

Hence,
E = \frac{2\lambda}{4\pi \epsilon_0 z}
 
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.
\varphi(z)=\int_{z}^{\infty}E(\zeta)d\zeta=\infty
 
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