Have a look at the diagram(adsbygoogle = window.adsbygoogle || []).push({});

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!

**Physics Forums - The Fusion of Science and Community**

# Electric Field due to a line of charge

Know someone interested in this topic? Share a link to this question via email,
Google+,
Twitter, or
Facebook

Have something to add?

- Similar discussions for: Electric Field due to a line of charge

Loading...

**Physics Forums - The Fusion of Science and Community**