• Support PF! Buy your school textbooks, materials and every day products Here!

[Gauss' Law] Hollow insulating sphere?

  • #1

Homework Statement



A hollow insulating sphere with an inner radius of 6.58 cm and outer radius of 11.7 cm has a uniform charge density of 79.9 μC/m3 distributed throughout the volume between.

If we want to use Gauss' Law to find the electric field at r = 17.2 cm, what "charge enclosed" should we use?

Homework Equations



RHO = Q/V

a = .0658 m
b = .1170 m
r = .1720 m

therefore r > b > a

The Attempt at a Solution



i kno RHO = Q/V

i just don't kno what V to use

i've used v = 4/3(pi)(.1720^3)

i've used v = (4/3)(pi)(.1720^3 - .1170^3)

think i've used every combination of r

i kno i'm overlooking something incredibly simple
blah!
 

Answers and Replies

  • #2
The charge is only in the middle region and the entire charge is enclosed in a sphere of radius r.

ie,v = (4/3)(pi)(.1170^3 - .0658^3)
 
  • #3
yes that makes perfect sense
and i could have swore i used those r's for v

thanks humanist
 

Related Threads on [Gauss' Law] Hollow insulating sphere?

  • Last Post
Replies
2
Views
5K
  • Last Post
Replies
6
Views
3K
Replies
2
Views
2K
Replies
1
Views
7K
Replies
9
Views
2K
Replies
1
Views
3K
Replies
6
Views
2K
Replies
1
Views
4K
Top