Help Solving Gauss' Law Homework Problem

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SUMMARY

The discussion focuses on solving a Gauss' Law problem involving an electric field inside a sphere with a volume charge density defined as ρ = k/r. The user successfully set up a spherical shell and calculated the enclosed charge using integration, resulting in E = k/(2ε). The user expresses confusion regarding the constancy of the electric field despite the variable charge density, attributing it to the relationship between the surface area of the Gaussian shell and the charge density's decline as r increases.

PREREQUISITES
  • Understanding of Gauss' Law and its applications in electrostatics
  • Familiarity with electric field concepts and charge density
  • Knowledge of integration techniques for calculating enclosed charge
  • Basic principles of spherical symmetry in physics
NEXT STEPS
  • Study the implications of charge density variations on electric fields
  • Learn about the derivation and applications of Gauss' Law in different geometries
  • Explore the concept of electric field continuity and its mathematical foundations
  • Investigate the behavior of electric fields in non-uniform charge distributions
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Students studying electromagnetism, physics educators, and anyone seeking to deepen their understanding of Gauss' Law and electric fields in varying charge distributions.

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Homework Statement


Trying to find E field inside a sphere radius a with volume charge density rho = k/r


Homework Equations



Gauss' law

The Attempt at a Solution



I set up a spherical shell radius R (R<a)
I found the charge inside by integrating rho from 0 to R (Q = 2*pi*a*R^2)

put these infos into gauss law, got E = k/(2(epsilon))

It just seems weird that the E field inside is constant even though the volume density is not. I guess it might be because the surface area of the gauss shell increases with R but the density drops off as 1/R.

Also, integrating from the origin seems like it might be a mistake (rho -> inf as r-> 0)
 
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Rho is the volume charge density, correct? Is there something you're missing when you go to find the charge enclosed?
 


Just a point of frustration on my part, please name your thread appropriately next time - I believe the rules do specify this.
 

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