Electric potential of concentr

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SUMMARY

The discussion focuses on calculating the electric potential of a cylindrical conductor placed within a cylindrical conducting shell. The inner cylinder has a charge density of λ0 = 2.60 µC/m, while the outer shell has a charge density of λ = -2.90 µC/m. The relevant radii are r0 = 8.40 cm, r1 = 16.80 cm, and r2 = 21.00 cm. The electric potential at r = 30.80 cm is determined to be -5.41E+03 V, with the solution involving the integration of the electric field E = 2kλ/r and the application of Gauss' theorem to simplify the calculations.

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Electric potential of cylindrical conductor placed in a cylindrical conducting shell

Homework Statement




dcylcon.gif
gcylcon.gif


A cylindrical conducting shell is placed concentric with a cylindrical conductor. Assume that a total charge density λ0 = 2.60 µC/ m is placed on the inner cylinder and a total charge density λ = -2.90 µC/ m is placed on the outer one, r0 = 8.40 cm, r1 = 16.80 cm and r2 = 21.00 cm. Calculate, relative to r = 10·r0 = 84.0 cm:

[1pt] the electric potential at r = 30.80 cm.


Answer: -5.41E+03 V
(I need help figuring out HOW to get this answer, not the answer itself)




Homework Equations




V= the negative inegral of E
λ=Q/2(pi)r


The Attempt at a Solution



So far all I've done is that I've integrated E=2kλ/r and got v=(-2kλ)ln(r)+C for the conducting shell and Vo=(-2kλo)ln(ro)+C for the conductor.

I'm not really sure what to do with the question, though.

I'm thinking I could use those equations to find the potential of both the cylindrical conductor and the cylindrical conducting shell. Then what?

And for the shell, would I have to use the two radiuses to find it on the outside of the shell and the inside too? I have no idea, I need assistance haha.

Help would be appreciated.
 
Last edited:
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Gauss' theorem will tell you that the E field depends only on the total charge density contained within the radius. In this case both of your points are outside of the outer shell, so none of the details of the inside matter. You can also do it by just summing the two potentials you have integrated and taking the difference between the two radii.
 
okay, I'll give it a shot, thanks!
 

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