Insulating sphere inside a hollow, conducting sphere (electric fields, Gauss's Law)

  • #1

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 > c


B). 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 rest


This 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.
 

Answers and Replies

  • #2
20
0


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:
  • #3
Office_Shredder
Staff Emeritus
Science Advisor
Gold Member
3,750
99


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
 
  • #4


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?
 

Related Threads on Insulating sphere inside a hollow, conducting sphere (electric fields, Gauss's Law)

Replies
2
Views
3K
Replies
6
Views
2K
Replies
1
Views
9K
  • Last Post
Replies
6
Views
3K
Replies
1
Views
2K
Replies
2
Views
5K
Replies
3
Views
3K
Replies
1
Views
4K
Replies
3
Views
26K
Top