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

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SUMMARY

The discussion focuses on deriving the radial charge distribution expression for an electric field (E field) using Gauss's law. Participants confirm that the divergence equation simplifies to ##\nabla \cdot \vec{E}=\frac{1}{r^2}\frac{\partial}{\partial{r}}(r^2 E_r)## due to the absence of E field components in the theta and phi directions. The final expression for charge density, ##\rho (r)##, is confirmed as ##\frac{2 E_o \epsilon_o}{r}##, demonstrating the relationship between electric field strength and charge distribution.

PREREQUISITES
  • Understanding of Gauss's Law in differential form
  • Familiarity with vector calculus, particularly the divergence operator
  • Knowledge of electric field concepts and notation
  • Basic understanding of charge density and its relationship to electric fields
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  • Study the application of Gauss's Law in various geometries
  • Learn about the divergence theorem and its implications in electromagnetism
  • Explore the relationship between electric fields and charge distributions in different materials
  • Investigate advanced topics in electrostatics, such as boundary conditions and potential theory
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Students and professionals in physics, electrical engineering, and anyone studying electromagnetism who seeks to understand the derivation of charge distributions from electric fields.

Jaccobtw
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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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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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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}##
 
Jaccobtw said:
Thank you. I got ##\frac{2 E_o \epsilon_o}{r}##
Looks good.
 
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