Question about charge density in a solid sphere

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SUMMARY

The discussion centers on calculating the surface charge density (σ) of a spherical shell within a uniformly charged solid sphere, characterized by total charge Q and radius R. The volume charge density (ρ) is defined as ρ = Q/(4/3 π R³). The participant seeks clarification on the expression σ = ρ dr, which relates the surface charge density to the volume charge density and an infinitesimal thickness dr of the shell. The participant correctly identifies that the charge dq in the shell is derived from the volume charge density and the volume element dV = 4πr²dr, leading to the conclusion that σ can be expressed as ρ(r)dr.

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  • Understanding of electrostatics and charge distributions
  • Familiarity with the concept of volume charge density
  • Knowledge of calculus, particularly integration and infinitesimals
  • Basic principles of spherical geometry
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  • Explore the concept of surface charge density in different geometries
  • Learn about the implications of charge density on electric potential
  • Investigate the relationship between volume charge density and electric field strength
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LostInToronto
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I've already posted this question in the advanced physics forum, but I really think it should go here. My apologies for the double posting.

Homework Statement



If we are given a uniformly charged solid sphere with total charge Q and radius R, then the volume charge density rho is given by

\rho = \frac{Q}{\frac{4}{3} \pi R^3}.

My question is: How do we express the surface charge density sigma of a spherical shell of infinitesimal width dr, located within this solid sphere?

Homework Equations





The Attempt at a Solution



I keep reading that

\sigma = \rho dr

but I really don't understand why this is the case. If someone could help clear this up, I'd really appreciate it.
 
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Ok, the charge dq contained in a shell of radius r and thickness dr is rho(r) times volume dV=4*pi*r^2*dr. Now surface charge density is charge dq divided by surface area 4*pi*r^2. So I suppose that would give you rho(r)*dr. But I'm not sure I'd really call that a 'surface charge'. I think it's really just a slice of the volume charge.
 
Thank you.
 

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