Help! Electric Field in Sphere with Uniform Charge

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SUMMARY

The discussion focuses on calculating the electric field in a solid insulating sphere with uniform charge density (ρ) and total charge (Q), surrounded by a conducting hollow sphere. The electric field is derived using Gauss's Law, resulting in E = Qr/(4πε0a³) for the region r < a. The confusion arises from the distinction between volume charge density and surface area in Gauss's Law. The importance of the inner sphere being an insulator is emphasized, as it allows for a uniform charge density without affecting the electric field inside.

PREREQUISITES
  • Understanding of Gauss's Law for electric fields
  • Familiarity with electric field concepts in conductors and insulators
  • Knowledge of charge density (ρ) and its implications
  • Basic calculus for volume and surface area calculations
NEXT STEPS
  • Study the derivation of electric fields using Gauss's Law in various geometries
  • Learn about the behavior of electric fields in conductors versus insulators
  • Explore the concept of induced charge and its calculation on conductors
  • Investigate the implications of charge density on electric field strength
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Students studying electromagnetism, physics educators, and anyone seeking to understand electric fields in charged spherical systems.

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Homework Statement


A solid, insulating sphere of radius a has a uniform charge density of ρ and a total charge of Q. Concentric with this sphere is an uncharged, conducting hollow sphere whose inner and outer radii are b and c, as shown.

p24-57.gif
A). Find the magnitude of the electric field in the following regions:

r < a (Use the following as necessary: ρ, ε0, and r.)

a < r < b

b < r < c

r > cB). Determine the induced charge per unit area on the inner and outer surfaces of the hollow sphere.

Homework Equations


Gauss's law for electric field (?):

E∫dA = (q in) / (ε0)

E = ke q / r2

E = 0 inside a CONDUCTOR

q = σdA (surface area?)

q = ρdV (volume? )

The Attempt at a Solution


I'm just concerned about the first part r< a and hopefully I will understand the restThis is just really tough for me... so many things I need to look out for and it is really confusing.

It is confusing when thinking about dimensions as the uniform charge is concerned with volume ( q = ρdV) yet I thought Gauss's Law ( flux = E ∫ dA ) was just concerned with 2D surface area only. Or am I missing something? Also I am not sure how the inner sphere, being an insulator, has any affect on the electric field.So...

for r < a, the electric field is not 0 and so

E ∫ dA = q in / ε0

E (4πr2) = q in / ε0

E = q in / (4πr2ε0)


which turns out wrong for r < a. It is still wrong even when I substitute q in as pV.
 
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Let me try

I think you were doing fine but substituted the wrong charge:

Q=ρV=ρ((4/3)πa3) therefor ρ=Q/((4/3)πa3)
for the new volume inside the sphere then q=ρ/((4/3)πr3)
if you substitute those to what you found then you get E = q in / (4πr2ε0) = ρ((4/3)πr3)/(4πr2ε0)
this equals ρr/3ε0 and if you substitute for ρ then Q/((4/3)πa3)r/3ε0 = Qr/(4/3)πa30 which equals Qr/4πε0a3
 
Last edited:


Also I am not sure how the inner sphere, being an insulator, has any affect on the electric field.

If, for example, the inner sphere was a conductor, it would be impossible to have a uniform charge density (as then the electric field inside the sphere would not be zero). This is the only reason they're stressing that it is an insulator
 


I got all the answers to it now, thanks for your help.

How would I go about finding the induced charges now? How would I start?
 

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