Electric Potential Of Charged Finite Rod

AI Thread Summary
The discussion focuses on calculating the electric potential at point P1 on the z-axis due to a uniformly charged finite rod, extending from z=-d to z=d. The potential is expressed using the integral V = k ∫ λ dz'/z, where the user seeks guidance on setting up the boundaries for the integral. The conversation highlights the importance of understanding the distances involved, specifically R, r, and r', in the context of the problem. A successful integration led to a potential result of λ*K*ln(3), but the user is reminded to also solve for an equal potential point P2 on the x-axis. The vector method discussed is noted as a useful approach for similar problems.
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Homework Statement


A thin rod extends along the z-axis from z=-d to z=d, carrying uniformly distributed charge along it's length with charge density lambda. Calculate the potential at P1 on the z-axis with coordinates (0,0,2d). Then find an equal potential at point P2 somewhere on the x-axis


Homework Equations




The Attempt at a Solution


I tried setting up an integral for the potential along the z-axis as such:
V= \int(1/4\pi\epsilon) * \lambdadz/z

but I'm not 100% sure how I should set up the boundaries? Also, since there aren't going to be any actual number answers I'm unsure as to how to equivocate my answer to some potential on the x-axis.

Thanks for any help.
 
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Well, the equation for V should actually read

V = k \int \lambda dz'/z

if that will help. Whenever a problem in EM is not immediately obvious to me, I always go to vectors.

V = k \int \lambda dl' \hat{R}/RIn this case you have R, which is the distance from the charge under consideration to the point of interest, r, which is the distance from the origin to the point of interest, and r', which is the distance from the origin to the charge under consideration. We know that R=r-r', and we need to sum up (integrate) over all the little possible charges of interest.

Give the boundaries and solution a try. You'll know you're on the right track if you get ln3.
 
I see, I integrated assuming P1 to be the origin, 0, and then from d to 3d and got lambda*K*ln(3) !

Thank You!
 
Yeah, that's the best way to do this particular set-up. You still have have to do the second part of the problem! :p

P.S. With that vector method I described to use its usually useful to note that (though in the Z part of the problem it didn't really come into play)

\hat{R}=\mathbf{R}/R
 
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