Charge density as a function of r

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SUMMARY

The discussion focuses on deriving the volume charge density (ρr) as a function of radius (r) for a sphere of radius R, where the electric field Er is defined as ER(r^4/R^4). Participants reference Gauss's Law, specifically the divergence form, \nabla\cdot E = \rho/\epsilon_0, and the relationship between charge (q) and volume. The confusion arises around the calculation of volume, as 4πr^2 represents the surface area of a sphere, not its volume.

PREREQUISITES
  • Understanding of Gauss's Law and its applications
  • Familiarity with electric fields and charge density concepts
  • Knowledge of calculus, particularly integration
  • Basic principles of electrostatics
NEXT STEPS
  • Review the derivation of volume charge density in electrostatics
  • Study the application of Gauss's Law in spherical coordinates
  • Learn about the relationship between surface area and volume in three-dimensional geometry
  • Explore advanced topics in electromagnetism, such as Maxwell's equations
USEFUL FOR

Students studying electromagnetism, physics educators, and anyone interested in understanding charge distribution in spherical geometries.

JJfortherear
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Homework Statement



Sphere of radius R, Er=ER(r4/R4)
Find the volume charge density (ρr) as a function of r

Homework Equations



Gauss's law

The Attempt at a Solution



I get that dQ=4πr2ρrdr, but have no idea how to solve for ρ
 
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Have you seen the following form of Gauss's Law?

[tex]\nabla\cdot E = \rho/\epsilon_0[/tex]
 
like kuruman said, its application of gauss's law

Flux = [tex]\oint[/tex]E (dot) dA = q/e

you found q to be 4πr2ρ, but p = Charge / Volume

but 4πr^2 isn't the volume of a sphere
 

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