Gauss's Law - A nonconducting spherical shell

In summary: P.S. You could also check by using ∫E dA = Q_enc/ε_0 to find the electric field inside a conducting sphere with uniform charge density ρ. Does it give the same answer as you got before?In summary, the conversation discussed using Gauss's law to derive an equation for the magnitude of the electric field at different radial distances from the center of a nonconducting spherical shell with uniform volume charge density. The equations for both a) and b) were derived using the equation ∫E dA = Q_enc/ε_0, with the gaussian surface being between the two shells for a) and outside both shells for b). However, the expression for a) may be incorrect and needs to be
  • #1
Edasaur
5
0
1. Homework Statement

A nonconducting spherical shell of inner radius R1 and outer radius R2 contains a uniform volume charge density ρ throughout the shell. Use Gauss's law to derive an equation for the magnitude of the electric field at the following radial distances r from the center of the sphere. Your answers should be in terms of ρ, R_1, R_2, r, ε_0, and π.

a) R_1 < r < R_2
b) r > R_2

2. Homework Equations
∫E dA = Q_enc/ε_0


3. The Attempt at a Solution

For a), I tried using Gauss's law to find it and I arrived at:

E = [ρ(R_1)^3]/[3(ε_0)(r^2)]

For b), I also used Gauss's law to find:

[ρ(R_1)^3 + ρ(R_2)^3]/[3(ε_0)(r^2)]


I'm not quite sure what I'm doing wrong...
 
Physics news on Phys.org
  • #2
Simply posting your answers is next to useless. Show your work.
 
  • #3
Ok. Here's my work:

For a), my gaussian surface is between the two shells
For b), my gaussian surface is outside both shells

a) ∫E dA = Q_enc/ε_0
E(4πr^2) = [ρ((4/3)π(R_1)^3)]/[ε_0]
E = [ρ((4/3)π(R_1)^3]/[(ε_0)(4πr^2)]
E = [ρ(R_1)^3]/[3(ε_0)(r^2)]

b)∫E dA = Q_enc/ε_0
E(4πr^2) = [ρ((4/3)π(R_1)^3) + ((4/3)π(R_2)^3)]/[ε_0]
E = [ρ(R_1)^3 + ρ(R_2)^3]/[3(ε_0)(r^2)]
 
  • #4
Edasaur said:
Ok. Here's my work:

For a), my gaussian surface is between the two shells
For b), my gaussian surface is outside both shells

a) ∫E dA = Q_enc/ε_0
E(4πr^2) = [ρ((4/3)π(R_1)^3)]/[ε_0]
E = [ρ((4/3)π(R_1)^3]/[(ε_0)(4πr^2)]
E = [ρ(R_1)^3]/[3(ε_0)(r^2)]
Shouldn't the amount of charge inside the sphere vary with r? The expression you have on the righthand side is a constant that's equal to the amount of charge in a solid sphere of radius R1 with charge density ρ. It's not applicable to this problem.

b)∫E dA = Q_enc/ε_0
E(4πr^2) = [ρ((4/3)π(R_1)^3) + ((4/3)π(R_2)^3)]/[ε_0]
E = [ρ(R_1)^3 + ρ(R_2)^3]/[3(ε_0)(r^2)]
I think if you figure out a), you'll see what's wrong here.
 
  • #5
I just did a) again:
∫E dA = Q_enc/ε_0
E(4πr^2) = [ρ((4/3)π(r/R_1)^3)]/[ε_0]
E = (ρr)/(3(ε_0)(R_1)^3)

Does that seem right?
 
  • #6
When r=R1, does it give the right answer?
 

1. What is Gauss's Law?

Gauss's Law is a fundamental law in electromagnetism that relates the electric flux through a closed surface to the enclosed electric charge. It is one of Maxwell's equations and is used to calculate the electric field at a point due to a distribution of charges.

2. What is a nonconducting spherical shell?

A nonconducting spherical shell is a spherical object made of a material that does not allow the flow of electric charges. This means that the charges on the surface of the shell cannot move freely, unlike in a conducting sphere where the charges can redistribute themselves.

3. What is the significance of Gauss's Law for a nonconducting spherical shell?

Gauss's Law for a nonconducting spherical shell states that the electric field inside the shell is zero. This is because the charges on the surface of the shell are fixed in place and do not contribute to the electric field inside the shell. This result is independent of the distribution of charges on the surface of the shell.

4. How is Gauss's Law applied to a nonconducting spherical shell?

To apply Gauss's Law to a nonconducting spherical shell, we first choose a Gaussian surface that encloses the shell. This surface can be a sphere with the same center as the shell. We then calculate the electric flux through this surface and equate it to the enclosed charge divided by the permittivity of free space. This allows us to calculate the electric field at any point inside or outside the shell.

5. What are some real-world applications of Gauss's Law for a nonconducting spherical shell?

Gauss's Law for a nonconducting spherical shell has many practical applications. For example, it can be used to calculate the electric field inside a charged insulating sphere, such as a balloon or a soap bubble. It is also used in electrostatic applications, such as in the design of capacitors, where nonconducting spherical shells are often used as the outer conductor. Additionally, Gauss's Law is used in theoretical and experimental studies of electromagnetism to understand the behavior of electric fields and charges.

Similar threads

  • Advanced Physics Homework Help
Replies
6
Views
1K
  • Advanced Physics Homework Help
Replies
9
Views
2K
  • Advanced Physics Homework Help
Replies
2
Views
2K
  • Introductory Physics Homework Help
Replies
23
Views
349
  • Advanced Physics Homework Help
Replies
10
Views
3K
  • Introductory Physics Homework Help
Replies
19
Views
1K
  • Introductory Physics Homework Help
Replies
2
Views
6K
  • Advanced Physics Homework Help
Replies
4
Views
3K
Replies
1
Views
871
Replies
4
Views
814
Back
Top