Derive an expression for the radial charge distribution of an E field

In summary, the conversation discusses the process of writing an expression for ##\rho (r)## using Gauss' law and the del equation. It is determined that the second and third terms of the del equation can be cancelled out due to the direction of the E field, leaving an expression of ##\nabla \cdot \vec{E}=\frac{1}{r^2}\frac{\partial}{\partial{r}}(r^2 E_r)##. The question asks for an expression of ##\rho (r)##, which can be achieved by isolating ##\rho## in the differential form of Gauss' law, resulting in ##\frac{\rho}{\epsilon_o}=\frac{2 E
  • #1
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Homework Statement
Consider a spherical distribution of charge that creates a uniform, radial electric field described by: ##\vec{E}=E_o\hat{r}##

Use the differential form of Gauss's Law to derive an expression for the radial charge distribution, ##\rho##, that will create this field. You will need the divergence in spherical coordinates: ##\nabla \cdot \vec{V}=\frac{1}{r^2}\frac{\partial}{\partial{r}}(r^2 V_r)+\frac{1}{r\sin[{\theta}]}\frac{\partial}{\partial{\theta}}(\sin[{\theta}]V_{\theta})+\frac{1}{r\sin[{\theta}]}\frac{\partial}{\partial{\phi}}(V_{\phi})##
Enter your mathematical expression for ρ(r) in terms of ##\epsilon_o, E_o ##, and ##r##
Relevant Equations
$$\nabla \cdot \vec{V}=\frac{1}{r^2}\frac{\partial}{\partial{r}}(r^2 V_r)+\frac{1}{r\sin[{\theta}]}\frac{\partial}{\partial{\theta}}(\sin[{\theta}]V_{\theta})+\frac{1}{r\sin[{\theta}]}\frac{\partial}{\partial{\phi}}(V_{\phi}) $$
I know we're supposed to attempt a solution but I'm honestly super confused here. I think the second an third terms of the del equation can be cancelled out because there is only an E field in the r hat direction, so no e field in the theta and phi directions. That leaves us with ##\nabla \cdot \vec{E}=\frac{1}{r^2}\frac{\partial}{\partial{r}}(r^2 E_r)##. The question says to write an expression for ##\rho (r)##. I know gauss' law has the differential form of itself equaling ##\frac{\rho}{\epsilon_o}##. would you isolate ##\rho## to get an expression for ##\rho (r)##? Thanks for your help
 
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  • #2
Jaccobtw said:
I think the second an third terms of the del equation can be cancelled out because there is only an E field in the r hat direction, so no e field in the theta and phi directions.
Yes.

Jaccobtw said:
That leaves us with ##\nabla \cdot \vec{E}=\frac{1}{r^2}\frac{\partial}{\partial{r}}(r^2 V_r)##.
The symbol ##V_r## on the right side would be better written as ##E_r##.

Jaccobtw said:
The question says to write an expression for ##\rho (r)##. I know gauss' law has the differential form of itself equaling ##\frac{\rho}{\epsilon_o}##. would you isolate ##\rho## to get an expression for ##\rho (r)##?
Yes. See what you get.
 
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  • #3
TSny said:
Yes.The symbol ##V_r## on the right side would be better written as ##E_r##.Yes. See what you get.
Thank you. I got ##\frac{2 E_o \epsilon_o}{r}##
 
  • #4
Jaccobtw said:
Thank you. I got ##\frac{2 E_o \epsilon_o}{r}##
Looks good.
 
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