Help! Electric Field in Sphere with Uniform Charge

[Color_of_Cyan]

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.

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.

 

[david13579]

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)πa³3ε₀ which equals Qr/4πε₀a³

 

[Office_Shredder]

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

 

[Color_of_Cyan]

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?

 

[asteriatic]

To solve this problem, you will need to use Gauss's Law for electric field, which states that the electric flux through a closed surface is equal to the enclosed charge divided by the permittivity of free space (ε0). In this case, the closed surface will be a spherical surface with radius r, centered at the center of the sphere.

For r < a:
Since the electric field inside a conductor is always zero, the electric field at any point inside the solid sphere (r < a) will be zero. This is because the conducting hollow sphere shields the electric field inside the solid sphere. Therefore, for this region, the electric field is 0.

For a < r < b:
To find the electric field in this region, we will use Gauss's Law. The Gaussian surface will be a spherical surface with radius r, centered at the center of the sphere. The enclosed charge will be the charge of the solid sphere, which is Q. Therefore, the electric field can be calculated as:

E ∫ dA = Q/ε0

E (4πr2) = Q/ε0

E = Q / (4πε0r2)

Substituting Q = ρ(4/3)πa3, we get:

E = ρa3 / (3ε0r2)

For b < r < c:
Similar to the previous case, we use Gauss's Law to calculate the electric field in this region. The Gaussian surface will be a spherical surface with radius r, centered at the center of the sphere. The enclosed charge will be the charge of the solid sphere, which is Q, minus the charge of the hollow sphere, which is 0 since it is uncharged. Therefore, the electric field can be calculated as:

E ∫ dA = Q/ε0

E (4πr2) = Q/ε0

E = Q / (4πε0r2)

Substituting Q = ρ(4/3)πb3, we get:

E = ρb3 / (3ε0r2)

For r > c:
In this region, the electric field is only due to the charge of the solid sphere, since the hollow sphere does not contribute to the electric field. Therefore, the electric field can be calculated as:

E ∫ dA = Q/ε0

E (4πr2) = Q/ε