Electric potential on an infinite line of charge

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Homework Help Overview

The problem involves calculating the electric potential at a specific radial distance from an infinite line charge placed within a conducting cylindrical shell. The charge density and dimensions of the shell are provided, along with a reference point for potential.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the application of the potential formula and the limits of integration. There is a focus on understanding the reference point for potential and the implications of the charge distribution on the potential calculation.

Discussion Status

Some participants are questioning the calculations and assumptions made regarding the reference point for potential. There is an acknowledgment of potential errors in the application of the formula, and a discussion on the correct limits for integration is ongoing.

Contextual Notes

Participants note that the potential cannot be set to zero at infinity due to the nature of the charge distribution, and there is a specific reference point chosen at the outer surface of the cylindrical shell.

Bryon
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Homework Statement


An infinite, uniform line charge with linear charge density λ = +5 µC/m is placed along the symmetry axis (z-axis) of an infinite, thick conducting cylindrical shell of inner radius a = 3 cm and outer radius b = 4 cm. The cylindrical shell has zero net charge.

The electrical potential is chosen to be zero at the outer surface of the cylindrical shell, V(b) = 0. (In this problem, it is not possible to chose the potential to be zero at infinity because the charge distribution extends to infinity.)

(c) Calculate the potential at a radial distance r = a/2.

?


Homework Equations



Vp-Vref = -∫E•dl = 2kλ*ln(Rref/R)

The Attempt at a Solution



Since the point b = 0.4 is the reference point I substituted that into the equation. Here is what I did:

R = 0.03/2 = 0.015

V = 2*(8.99x10^9)*(5x10^-6)*ln(0.04/0.015) = 88176.54985V

What did I miss?
 
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For a < r < b, E(r) = 0 .

∴ [tex]\int_{r=b}^{r}\vec{E}\cdot \vec{d\ell}=\int_{r=b}^{r=a}\vec{0}\cdot\vec{d\ell}+\int_{r=a}^{r}\vec{E}(r)\cdot\vec{d\ell}[/tex]

The non-zero part of the integral begins at r=a.
 
I understand where the potential is zero but I cannot figure out the potential at r=a/2. Did I calculate it wrong?
 
Bryon said:
I understand where the potential is zero but I cannot figure out the potential at r=a/2. Did I calculate it wrong?
Yes, it's wrong.

I showed where your "formula": Vp-Vref = -∫E•dl = 2kλ*ln(Rref/R) comes from.

So, it's as if Rref = a, not b. Technically it's because V(b) = V(a).
 

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