# Electric Field due to a line of charge

by stunner5000pt
Tags: charge, electric, field, line
 P: 1,439 Have a look at the diagram Find the elctric field a distance z above one end of a astraight line segment of lenght L which carries uniform line charge lambda. Check that your formula is consistent with what you would expect for the case z >> L SOlution: $$\lambda= q/L$$ $$dq = \lambda dx$$ For the electric Field in the horizontal (points to the left and is negative) $$dE_{x} = \frac{1}{4 \pi \epsilon_{0}} \frac{dq}{(z^2 + x^2)} \sin \theta$$ Subsituting what we know about sin theta and dq $$dE_{x} = \frac{1}{4 \pi \epsilon_{0}} \frac{\lambda dx}{(z^2 + x^2)} \frac{x}{\sqrt{z^2 + x^2}}$$ integrating x = 0 to x = L $$E_{x} = \int dE_{x} = \frac{\lambda}{4 \pi \epsilon_{0}} \int_{x=0}^{x=L} \frac{xdx}{(z^2+x^2)^{\frac{3}{2}}} = \frac{\lambda}{4 \pi \epsilon_{0}} \left[ \frac{-1}{\sqrt{z^2 + x^2}} \right]_{x=0}^{x=L} = \frac{\lambda}{4 \pi \epsilon_{0}} \left[ \frac{-1}{\sqrt{z^2 + L^2}} + \frac{1}{z} \right]$$ $$E_{x} = \frac{\lambda}{4 \pi \epsilon_{0}} \left[ \frac{1}{z} -\frac{1}{\sqrt{z^2 + L^2}} \right]$$ ok so suppose z >> L then the electric field is zero?? Shouldnt it reduce to that of a point charge and not zero?? this should be regardless of whether i solve for x or z right?? for the Z direction i got $$E_{z} = \frac{\lambda}{4 \pi \epsilon_{0}z} \frac{L}{\sqrt{L^2 + z^2}}$$ unlike the last one this one does reduce to the equatiopn for a point charge but is off my a factor of 1/z ... combining the two yields $$\vec{E} = \frac{\lambda}{4 \pi \epsilon_{0}} \left[ \frac{1}{z} -\frac{1}{\sqrt{z^2 + L^2}} \right] \hat{x} + \frac{\lambda}{4 \pi \epsilon_{0}z} \frac{L}{\sqrt{L^2 + z^2}} \hat{z}$$ is there soemthing wrong with the calculationg ofr hte X Horizontal Direction?? Please help Thank you in advance for your help and advice! Attached Thumbnails
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P: 41,438
 Quote by stunner5000pt $$E_{x} = \frac{\lambda}{4 \pi \epsilon_{0}} \left[ \frac{1}{z} -\frac{1}{\sqrt{z^2 + L^2}} \right]$$ ok so suppose z >> L then the electric field is zero?? Shouldnt it reduce to that of a point charge and not zero?? this should be regardless of whether i solve for x or z right??
If you get far enough away, the field from anything (finite) goes to zero! But you need to understand how it approaches zero. In this case, think about it: You are very far from a point charge--What's the direction of the field?

 for the Z direction i got $$E_{z} = \frac{\lambda}{4 \pi \epsilon_{0}z} \frac{L}{\sqrt{L^2 + z^2}}$$ unlike the last one this one does reduce to the equatiopn for a point charge but is off my a factor of 1/z ...
It's not off. You just need to examine its behavior as z goes to infinity. Use a binomial expansion:
$$(L^2 + z^2)^{-1/2} \approx (1/z)(1 - \frac{L^2}{2z^2})$$
 P: 3 I'm studying from Griffiths too :) and i've a question about this exercize. The electric potential: $$V = \frac{ \lambda }{4\epsilon_0 \pi} ln\left(\frac{L+\sqrt{z^2+L^2}}{z}\right)$$ Calculating the electric field from the V's gradient, it has x component always at zero. Where the problem? I'm sorry for my poor english, good evening mate!
P: 26
Electric Field due to a line of charge

Yeah, Griffith's book is good.

 Quote by stunner5000pt combining the two yields $$\vec{E} = \frac{\lambda}{4 \pi \epsilon_{0}} \left[ \frac{1}{z} -\frac{1}{\sqrt{z^2 + L^2}} \right] \hat{x} + \frac{\lambda}{4 \pi \epsilon_{0}z} \frac{L}{\sqrt{L^2 + z^2}} \hat{z}$$ is there soemthing wrong with the calculationg ofr hte X Horizontal Direction?? Please help Thank you in advance for your help and advice!
From your calculation for $$\vec{E}$$ (yes i think it's correct for both direction), for z >> L, the electric field of component - x will dissapear (since $$\frac{1}{z} - \frac{1}{z} = 0)$$ and leaves for us the z - component as below:

after simplifying, $$\vec{E} = \frac{\lambda L}{4 \pi \epsilon_{0}z^2}\hat{z}$$ for z>>L.

the case is the same as above example from griffith's book (that is for E at a distance z above midpoint of line L). I quoted it : From far away the line

"looks" like a point charge $$q = 2\lambda L$$.

 Quote by Peppe I'm studying from Griffiths too :) and i've a question about this exercize. The electric potential: $$V = \frac{ \lambda }{4\epsilon_0 \pi} ln\left(\frac{L+\sqrt{z^2+L^2}}{z}\right)$$ Calculating the electric field from the V's gradient, it has x component always at zero. Where the problem?
I think x and y component will be gone since you do grad $$\nabla$$-operation to potential (the potential only depend on z). so,

$$\vec{E} = -\nabla V = -\frac{\lambda}{4\pi \epsilon_{0}}\frac{\partial}{\partial z}\ln (\frac{L + \sqrt{z^2 + L^2}}{z})$$.

suppose $$k = \sqrt{z^2 + L^2}$$,

it will give $$\vec{E} = \frac{\lambda}{4 \pi \epsilon_{0}}(\frac{z}{k(L + k)} - \frac{1}{z})\hat{z}$$

btw, i cant find this problem in griffith's book....

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