Electric field around a sphere with an internal charge distribution

  • Thread starter goohu
  • Start date
  • #1
35
3

Homework Statement:

see picture

Relevant Equations:

integral E ds = integral charge distribution * 4*pi*R^2 dR
Untitled.png


I'm just going to skip some of the step since I only need help with understanding the last part.
After rearranging the equation stated at "Relevant equation" (and skipping some steps) we will get:
E * 4*pi*e0*R^2 = integral pv * 4*pi*R^2 dR

E = 1/(4*pi*e0*R^2) * 4*pi * integral pv*R^2 dR

E = 1/(e0*R^2) * Q/(pi*a^4) * integral R^3 dR

E = Q / (pi*a^4*e0*R^2) * integral R^3 dR

Now for the last step is where I'm unsure where to put the integral limits.

Inside the sphere (0<R<a): the integral should be from 0 to a?

Outside the sphere (R>a): the integral should be from a to R?
 

Attachments

Answers and Replies

  • #2
vela
Staff Emeritus
Science Advisor
Homework Helper
Education Advisor
14,845
1,416
Explain what Gauss's Law says. Be specific.
 
  • #3
Orodruin
Staff Emeritus
Science Advisor
Homework Helper
Insights Author
Gold Member
16,829
6,652
Homework Statement:: see picture
Relevant Equations:: integral E ds = integral charge distribution * 4*pi*R^2 dR

After rearranging the equation stated at "Relevant equation" (and skipping some steps) we will get:
E * 4*pi*e0*R^2 = integral pv * 4*pi*R^2 dR
We most certainly will not. Your left-hand side depends on R while it is an integration variable on the right.
 
  • Like
Likes SammyS
  • #4
35
3
Sorry for the late reply.

Gauss law states that the net electric flux of a closed surface is equal to the total electric charge inside the surface divided by e0.

The electric flux is the electric field vectors going through the surface at the normal angle.

E * 4*pi*r^2 = (1/e0) * ∫ pv dV , where pv is the charge distribution and dV is a volume unit.

We can rewrite dV into dr dtheta dphi-> looking at my formula sheet this becomes 4*pi*r^2 dr. This is the same answer to what I got before.

I don't understand what is wrong, there was a similar task solved in class by the teacher. That task was : "Determine the electric field everywhere from the charge distribution given by: " pv = {4R, 0<=R<1 ; 5^2 , 1<= R < 2 ; -8, R = 2}

I'm really confused.
 
  • #5
vela
Staff Emeritus
Science Advisor
Homework Helper
Education Advisor
14,845
1,416
E * 4*pi*r^2 = (1/e0) * ∫ pv dV , where pv is the charge distribution and dV is a volume unit.

We can rewrite dV into dr dtheta dphi-> looking at my formula sheet this becomes 4*pi*r^2 dr. This is the same answer to what I got before.
OK, so you're using a spherical Gaussian surface of radius ##r##. What limits do you use to find the charge inside the Gaussian surface for the case where ##r<a##?
 
  • #6
35
3
If its inside the sphere then I think the limits should be from 0 to r.

And what happens outside the sphere? I suppose outside the sphere is just empty space so no contribution to the effect on electric field. But it is still affected by the sphere. So what is the electric field outside the sphere?

Maybe integrate from 0 to a and then a to r?
 
  • #7
vela
Staff Emeritus
Science Advisor
Homework Helper
Education Advisor
14,845
1,416
If its inside the sphere then I think the limits should be from 0 to r.
Right.

And what happens outside the sphere? I suppose outside the sphere is just empty space so no contribution to the effect on electric field. But it is still affected by the sphere. So what is the electric field outside the sphere?

Maybe integrate from 0 to a and then a to r?
Good. Regardless of whether ##r<a## or ##r>a##, you still integrate from 0 to ##r## because you have to integrate over the entire volume enclosed in the Gaussian surface. When you're outside the sphere of charge, you get
$$E(4\pi r^2) = \int_0^r \rho(r') 4\pi r'^2\,dr = \int_0^a \rho(r') 4\pi r'^2\,dr' + \int_a^r \rho(r') 4\pi r'^2\,dr'.$$ But remember the charge density ##\rho## vanishes for ##r'>a##, so the last integral is zero.
 
  • Informative
Likes goohu

Related Threads on Electric field around a sphere with an internal charge distribution

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