Finding the Electric Field Outside of a Spherical Shell Using Gauss's Law

[skate_nerd]

Homework Statement

The problem states that you have a spherical shell with inner radius R_i=1 cm and outer radius R₀=2 cm. The shell also has uniform charge density of ρ=10^-3 N/m³. I found the first few answers of the question already. First was to get the charge of the shell, which is simply ρV_shell, or Q=ρ(4∏/3)(R₀³-R_i³). This ends up being 2.93(10^-8)C.
Next I found that the electric field magnitude everywhere inside the shell could be expressed in terms of r by just setting EA=ρ(4∏/3)(r³-R_i³)/ε₀, where the 4∏ cancels. the answer ends up being ((1/3)ρ/ε₀)(r-(R_i³/r²), or 3.76(10⁷) N/Cm.
Now the part I'm having trouble with is finding an equation in terms of r for the electric field magnitude outside of the spherical shell.

Homework Equations

Gauss's Law. Surface area and volume of a sphere.

The Attempt at a Solution

I tried using Gauss's law in a similar way to the last part of the problem by having EA=Q/ε₀, but I don't understand how you are supposed to solve for this if you don't know what the uniform charge density is outside of the shell. I don't feel like using that same value would make sense. Could somebody explain this to me?

 

[haruspex]

Next I found that the electric field magnitude everywhere inside the shell could be expressed in terms of r by just setting EA=ρ(4∏/3)(r³-R_i³)/ε₀, where the 4∏ cancels. the answer ends up being ((1/3)ρ/ε₀)(r-(R_i³/r²), or 3.76(10⁷) N/Cm.

How did you get that? There should not be a field inside an empty uniformly charged spherical shell.

 

[skate_nerd]

I said that is the E field inside the shell, as in like in the volume of the shell. Thats why I used the gaussian radius as arbitrary r minus the inner radius of the sphere.

 

[haruspex]

I said that is the E field inside the shell, as in like in the volume of the shell. Thats why I used the gaussian radius as arbitrary r minus the inner radius of the sphere.

OK - that's often a tricky distinction to make verbally.
The field outside a uniformly charged spherical shell happens to be exactly as though all the charge were concentrated at the sphere's centre.

I don't feel like using that same value would make sense.

Can you explain that some more? What same value?

 

[Nickie]

I would approach this problem by first considering the symmetry of the spherical shell and how it affects the electric field outside of the shell. Since the shell has uniform charge density and is symmetric, we can apply Gauss's law to find the electric field outside of the shell.

First, we need to choose a Gaussian surface that is symmetrical to the shell. In this case, a spherical Gaussian surface with radius r > R0 would be appropriate. This is because the electric field outside of the shell is affected by the total charge of the shell, which is contained within the Gaussian surface.

Next, we can use Gauss's law, which states that the electric flux through any closed surface is equal to the enclosed charge divided by the permittivity of free space (ε0). In this case, the enclosed charge is simply the charge of the shell, which we already calculated in the problem statement (Q = 2.93(10-8)C). Therefore, we can write:

EA(4∏r2) = Q/ε0

Solving for EA, we get:

EA = Q/(4∏ε0r2)

Substituting in the value of Q, we get:

EA = (2.93(10-8)C)/(4∏ε0r2)

This is the equation for the electric field magnitude outside of the spherical shell. It is important to note that this equation is valid for r > R0, as we chose a Gaussian surface with r > R0. For r < R0, the electric field magnitude is given by the equation you found in your attempt at a solution.

In summary, by considering the symmetry of the spherical shell and applying Gauss's law, we were able to find an equation for the electric field magnitude outside of the shell. As a scientist, it is important to always consider the symmetry of a system when solving problems, as it can often simplify the solution process.