Electric field of an infinitely long wire with radius R

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The discussion revolves around calculating the electric field and potential of an infinitely long wire with radius R. The user initially calculated the electric field as E = (ρ₀ R²)/(2 ϱ ε₀) and the potential as φ = (ρ₀ R²)/(2 ε₀) ln(ϱ/R), but faced discrepancies with the Poisson equation results. Clarifications were made regarding the use of cgs units, which omit ε₀, and the need to apply the correct form of the Laplacian in cylindrical coordinates. The user confirmed that the calculations for the electric field and potential inside the wire were completed, leading to a consistent result of 4πρ₀ in cgs units. Overall, the conversation highlights the importance of unit systems and proper coordinate transformations in electrostatics.
Lambda96
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Homework Statement
Calculate the potential and check if it satisfies the poisson equation.
Relevant Equations
none
Hi,

I don't know if I have calculated the electric field correctly in task a, because I get different values for the Poisson equation from task b

Bildschirmfoto 2024-10-29 um 21.42.42.png

The flow of the electric field only passes through the lateral surface, so ##A=2\pi \varrho L## I calculated the enclosed charge as follows ##q=\rho_0 \pi R^2 L## then I obtained the following electric field ##E=\frac{\rho_0 R^2}{2 \varrho \epsilon_0 }##

For the potential I got the following ##\phi=\frac{\rho_0 R^2}{2 \epsilon_0} \ln{\frac{\varrho }{R}}##

For the Poisson equation I then get ##-\Delta \phi=\frac{\rho_0 R^2}{2 \epsilon_0 \varrho^2 }##

Unfortunately, I don't know what I've done wrong, which also makes me wonder why the task only says ##4 \pi \rho## without the ##\epsilon_0##
 
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Did you use the right form for the Laplacian? Remember you're working in cylindrical coordinates.

For the problem as a whole, you also need to consider the region inside the wire.
 
Lambda96 said:
. . . which also makes me wonder why the task only says ##4 \pi \rho## without the ##\epsilon_0##
That's probably because the textbook where you found this is using cgs units. What does the expression for Coulomb's law look like?
 
vela said:
For the problem as a whole, you also need to consider the region inside the wire.
I thought about that but the problem in part (b) asks about the potential in the region outside the wire as deduced from the limits of integration. Of course, the Poisson equation is trivially satisfied in that region.

Screen Shot 2024-10-29 at 9.38.19 PM.png
 
Thank you vela and kuruman for your help 👍👍

Since the potential depends only on the radius, the Laplacian in cylindrical coordinates has the following form ##\frac{1}{\varrho} \frac{\partial}{\partial \varrho} \Bigl( \varrho \frac{\partial f}{\partial \varrho} \Bigr)##

Because of the limits for the integral, I also assumed that only the electric field and potential outside the wire is required

I then applied the Laplacian to the potential and got the following:

$$\frac{1}{\varrho} \frac{\partial}{\partial \varrho} \Bigl( \varrho \frac{\partial f}{\partial \varrho} \Bigr) \frac{\rho_0 R^2}{2 \epsilon_0} \ln{\frac{\varrho}{R}}=\frac{1}{\varrho} \frac{\partial}{\partial \varrho} \frac{\rho_0 R^2}{2 \epsilon_0}=0$$

Can it be that ##-\Delta \phi=4 \pi \rho## is valid for the inside of the wire?
 
I have now completed the calculation for the electric field and the potential within the wire. Then calculated the Laplacian and converted the result from SI to cgs and have now got ##4 \pi \rho_0##
 
I want to find the solution to the integral ##\theta = \int_0^{\theta}\frac{du}{\sqrt{(c-u^2 +2u^3)}}## I can see that ##\frac{d^2u}{d\theta^2} = A +Bu+Cu^2## is a Weierstrass elliptic function, which can be generated from ##\Large(\normalsize\frac{du}{d\theta}\Large)\normalsize^2 = c-u^2 +2u^3## (A = 0, B=-1, C=3) So does this make my integral an elliptic integral? I haven't been able to find a table of integrals anywhere which contains an integral of this form so I'm a bit stuck. TerryW

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