Electric field of spherical shell

Click For Summary
SUMMARY

The discussion focuses on calculating the electric field of a spherical shell with uniform charge density ρ, specifically addressing the regions outside (r > R2), between (R1 < r < R2), and inside (r < R1) the shell. The correct approach involves using the volume to determine the total charge, as indicated by the density ρ, rather than the shell area. The formula for the electric field outside the shell is derived as E = ρ(R2^3 - R1^3) / (3ε0r^2), confirming that the charge is distributed throughout the volume between the inner and outer radii of the shell.

PREREQUISITES
  • Understanding of electric fields and Gauss's law
  • Familiarity with spherical coordinates and volume calculations
  • Knowledge of charge density concepts
  • Basic calculus for integration and volume determination
NEXT STEPS
  • Study Gauss's law applications in electrostatics
  • Learn about electric fields of different charge distributions
  • Explore the concept of charge density in various geometries
  • Review calculus techniques for volume integration
USEFUL FOR

Students studying electromagnetism, physics educators, and anyone seeking to understand electric fields in spherical charge distributions.

starstruck_
Messages
185
Reaction score
8

Homework Statement


Consider a spherical shell with uniform charge density ρ.
The shell is drawn as a donut with inner (R1) and outer (R2) radii.
Let r measure the distance from the center of the spherical shell, what is the electric field at r>R2, R1<r<R2, and r<R1.

I am working on the r > R2 part right now.
I'll work on R1<r<R2 later.

The question says spherical shell so I would have assumed that we would multiply ρ by the area of the shell instead of the volume, however my professor's solution uses volume for the total charge. I solved the problem, I just don't understand why we would use volume and not the shell area. Although, it DOES give ρ which is the density per unit of volume.
It just doesn't make sense to me. What am I missing? Am I misunderstanding something?

Homework Equations


E = Qenclosed/ (ε0A)

The Attempt at a Solution


For r > R2 [/B]
Qenclosed = ρ*4/3π(R2^3-R1^3)
A = 4πr^2
R2 is the outer radius of the shell and r is the distance from the center.

E = ρ(R2^3-R1^3)/(3ε0*r^2)
I already know that r<R1 =0 from the center of the shell to R1, there is no charge (it is a hole).
 
Last edited:
Physics news on Phys.org
The charge is spread throughout a volume. Imagine two spheres with the same center. One sphere has radius R1 and the other has radius R2. The charge is spread throughout the volume between the surfaces of the two spheres. This region between the two spherical surfaces constitutes the "spherical shell".

See figure (a) here
https://www.chegg.com/homework-help/questions-and-answers/figure-b-nonconducting-spherical-shell-inner-radius-206-cm-outer-radius-b-256-cm-volume-ch-q3674386
 
TSny said:
The charge is spread throughout a volume. Imagine two spheres with the same center. One sphere has radius R1 and the other has radius R2. The charge is spread throughout the volume between the surfaces of the two spheres. This region between the two spherical surfaces constitutes the "spherical shell".

See figure (a) here
https://www.chegg.com/homework-help/questions-and-answers/figure-b-nonconducting-spherical-shell-inner-radius-206-cm-outer-radius-b-256-cm-volume-ch-q3674386
AHH right! That makes sense! Thank you (totally forgot about that from calculus)
 

Similar threads

Replies
5
Views
1K
Replies
1
Views
943
  • · Replies 5 ·
Replies
5
Views
748
  • · Replies 17 ·
Replies
17
Views
2K
  • · Replies 7 ·
Replies
7
Views
5K
Replies
4
Views
2K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 7 ·
Replies
7
Views
1K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 4 ·
Replies
4
Views
4K