Help! Electric Field in Sphere with Uniform Charge

Click For Summary

Homework Help Overview

The problem involves a solid insulating sphere with a uniform charge density and a concentric uncharged conducting hollow sphere. Participants are tasked with finding the electric field in various regions and determining the induced charge on the surfaces of the hollow sphere.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster expresses confusion regarding the application of Gauss's law and the relationship between volume and surface area in the context of electric fields. They attempt to calculate the electric field for the region inside the insulating sphere but question their approach.
  • Another participant suggests a correction to the charge substitution in the calculations, providing an alternative expression for the electric field.
  • There is a discussion about the implications of the inner sphere being an insulator versus a conductor, with questions raised about how this affects the electric field.

Discussion Status

Participants are actively engaging with the problem, offering corrections and clarifications. The original poster has indicated progress in understanding the electric field calculations but is seeking further guidance on determining induced charges.

Contextual Notes

Participants are navigating the complexities of applying Gauss's law in different regions and the implications of charge distribution in insulating versus conducting materials. There is an acknowledgment of the challenges posed by the dimensional aspects of the problem.

Color_of_Cyan
Messages
386
Reaction score
0

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.
 
Physics news on Phys.org


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?
 

Similar threads

Replies
6
Views
2K
Replies
23
Views
5K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 11 ·
Replies
11
Views
2K
Replies
5
Views
1K
  • · Replies 5 ·
Replies
5
Views
6K
  • · Replies 18 ·
Replies
18
Views
5K
  • · Replies 8 ·
Replies
8
Views
3K
  • · Replies 7 ·
Replies
7
Views
2K