Gauss' Law question

  • #1

Homework Statement


a spherical shell has an outer radius R and a inner radius R/2 and carries a total charge -q, distributed with uniform charge density. A point charge +q is at the centre of the sphere. Calculate the electric field strength for R/2<r<R

Homework Equations



Gauss' Law

The Attempt at a Solution



I figured out the charge density in the shell ρ = [itex]\frac{-6q}{7πR^3}[/itex], so the charge enclosed by a gaussian sphere is q(1-[itex]\frac{8r^3}{7R^3}[/itex])

Then using gauss' law I get E = [itex]\frac{q}{4πε_0r^2}[/itex](1-[itex]\frac{8r^3}{7R^3}[/itex])

But my book says E = [itex]\frac{q}{4πε_0r^2}[/itex][itex]\frac{8}{7}[/itex](1-[itex]\frac{r^3}{R^3}[/itex])


Also does anyone know why the latex things aren't working? I'm new to all this stuff.

Edit: cheers guys
 
Last edited:

Answers and Replies

  • #2
760
69
Also does anyone know why the latex things aren't working? I'm new to all this stuff.

Don't use the BB code tags like inside the LaTeX code. If you want to do superscripts in the LaTex math code, use ^ .
 
  • #3
Doc Al
Mentor
44,946
1,216
Also does anyone know why the latex things aren't working? I'm new to all this stuff.
Could be that you're combining Latex with non-Latex stuff. Stick to pure Latex and it should work.
 
  • #4
Doc Al
Mentor
44,946
1,216
I figured out the charge density in the shell ρ = [itex]\frac{-6q}{7πR^3}[/itex],
Good.

so the charge enclosed by a gaussian sphere is q(1-[itex]\frac{8r^3}{7R^3}[/itex])
Redo that one.

(FYI: I agree with the book's answer.)
 
  • #6
Doc Al
Mentor
44,946
1,216
how do you get
What's the definition of charge density?
 
  • #7
3
0
What's the definition of charge density?
i dont really understand charge density. all i know is that it have 3 charge densities, volume, area, and linear.
i just learned this today and i get ρ = −6q/πR3 not ρ = −6q/7πR3.
can you help me solve this step by step?
 
  • #8
Doc Al
Mentor
44,946
1,216
i dont really understand charge density. all i know is that it have 3 charge densities, volume, area, and linear.
Here we are talking about ρ, which is a volume density.

i just learned this today and i get ρ = −6q/πR3 not ρ = −6q/7πR3.
can you help me solve this step by step?
Why don't you show how you arrived at your answer?
 

Related Threads on Gauss' Law question

  • Last Post
Replies
2
Views
6K
  • Last Post
Replies
15
Views
2K
  • Last Post
Replies
13
Views
4K
  • Last Post
Replies
6
Views
1K
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
9
Views
1K
  • Last Post
Replies
7
Views
2K
  • Last Post
Replies
0
Views
2K
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
4
Views
6K
Top