Electric Field of Concentric Spheres and Opposite Charges

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SUMMARY

The discussion focuses on calculating the electric field between two concentric spheres with differing charge densities, specifically using Gauss' Law. The inner sphere has a positive charge density of +ρ, while the region between the spheres has a negative charge density of -ρ. Participants emphasize that the electric field in the region between the spheres (R1 < r < R2) can be determined using Gauss' Law, which simplifies to the electric field of a point charge due to the spherical symmetry of the charge distribution.

PREREQUISITES
  • Understanding of Gauss' Law in electrostatics
  • Familiarity with electric field concepts and charge density
  • Knowledge of spherical coordinates and symmetry in physics
  • Basic calculus for integrating over spherical surfaces
NEXT STEPS
  • Study the application of Gauss' Law in various geometries
  • Learn about electric fields generated by point charges and spherical charge distributions
  • Explore the concept of charge density and its effects on electric fields
  • Investigate the implications of spherical symmetry in electrostatics
USEFUL FOR

Students of physics, particularly those studying electromagnetism, as well as educators and anyone seeking to deepen their understanding of electric fields and charge distributions in concentric spherical systems.

idkwhatimdoing
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Homework Statement


The figure to the right shows two concentric spheres made from insulators. One has radius and the other has radius R1, and the other has radius R2. The inner sphere has a positive charge density, +ρ, while the insulator region between the inner and outer spheres has a negative charge density, −ρ.

IWL71Lh.png

Homework Equations


Gauss' Law

The Attempt at a Solution


a. I wrote expressions that represented each portion's charge densities, and set them equal to each other because the magnitudes of the charge densities are the same; I got 2.

b. I used a Gaussian sphere with radius r and enclosed a portion of the smaller sphere; solved

c. I used a Gaussian sphere with radius R2>r>R1. I expressed the charge of the small sphere as (4/3)π(R1)ρ. The other portion would be a sphere with radius r with a spherical cavity with radius R1 and charge density -ρ. I expressed its charge as (4/3)π(r^3-R1^3)(-ρ). I added these two together to get the enclosed charge, but I'm not sure how to express the flux in terms of E and dA. I know the negatively charged portion would have field lines entering the surface, and thus leading to a negative dot product, but would the positively charged sphere have an effect as well? Not sure what to do from here.

Any help would be appreciated
 
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You seem to be almost there. Having found the net charge inside the sphere radius r, R1<r<R2, isn't writing down the field at r immediate?
 
Would it be the same as the electric field of a point charge?
 
The formulation of this problem is terrible.

There is one sphere, the one with R = R1, then there is a spherical shell from r = R1 to R = R2. There are not two shells, nor an "insulating layer" between them.

Anyway, you're well on your way. I also don't see why it isn't immedialtely apparent to you what the E field has to be for R1 < r < R2. Just use Gauss's law! ε∫∫E⋅dA = Qfree inside the surface! It doesn't matter that some of the charge inside is + and some of it -, as long as all charge is distributed symmetrically with respect to the spherical coordinates φ and θ, which they are.
 
Last edited:
idkwhatimdoing said:
Would it be the same as the electric field of a point charge?
Yes, that's the neat thing about charges with a spherically symmetric distribution.
 

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