Electric field of an infinitely long wire with radius R

[Lambda96]

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

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 $\epsilon_0$

 

[vela]

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.

 

[kuruman]

. . . 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?

 

[kuruman]

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.

 

[Lambda96]

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?