
#1
Sep2106, 09:22 PM

P: 1,445

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: [tex] \lambda= q/L [/tex] [tex] dq = \lambda dx [/tex] For the electric Field in the horizontal (points to the left and is negative) [tex] dE_{x} = \frac{1}{4 \pi \epsilon_{0}} \frac{dq}{(z^2 + x^2)} \sin \theta [/tex] Subsituting what we know about sin theta and dq [tex] dE_{x} = \frac{1}{4 \pi \epsilon_{0}} \frac{\lambda dx}{(z^2 + x^2)} \frac{x}{\sqrt{z^2 + x^2}} [/tex] integrating x = 0 to x = L [tex] 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] [/tex] [tex] E_{x} = \frac{\lambda}{4 \pi \epsilon_{0}} \left[ \frac{1}{z} \frac{1}{\sqrt{z^2 + L^2}} \right] [/tex] 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 [tex] E_{z} = \frac{\lambda}{4 \pi \epsilon_{0}z} \frac{L}{\sqrt{L^2 + z^2}} [/tex] 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 [tex] \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} [/tex] is there soemthing wrong with the calculationg ofr hte X Horizontal Direction?? Please help Thank you in advance for your help and advice! 



#2
Sep2206, 07:26 AM

Mentor
P: 40,907

[tex](L^2 + z^2)^{1/2} \approx (1/z)(1  \frac{L^2}{2z^2})[/tex] 



#3
May1011, 02:15 PM

P: 3

I'm studying from Griffiths too :) and i've a question about this exercize.
The electric potential: [tex]V = \frac{ \lambda }{4\epsilon_0 \pi} ln\left(\frac{L+\sqrt{z^2+L^2}}{z}\right) [/tex] 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! 



#4
May1311, 05:40 AM

P: 26

Electric Field due to a line of charge
Yeah, Griffith's book is good.
after simplifying, [tex]\vec{E} = \frac{\lambda L}{4 \pi \epsilon_{0}z^2}\hat{z}[/tex] 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 [tex]q = 2\lambda L[/tex]. [tex]\vec{E} = \nabla V = \frac{\lambda}{4\pi \epsilon_{0}}\frac{\partial}{\partial z}\ln (\frac{L + \sqrt{z^2 + L^2}}{z})[/tex]. suppose [tex]k = \sqrt{z^2 + L^2}[/tex], it will give [tex]\vec{E} = \frac{\lambda}{4 \pi \epsilon_{0}}(\frac{z}{k(L + k)}  \frac{1}{z})\hat{z}[/tex] btw, i cant find this problem in griffith's book.... 


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