# Thick spherical shell confirmation of solution

• Liquidxlax
In summary: The charge enclosed is Qenc = Q, so Gauss' law simplifies to E = kQ/r^2, which is the same equation as the point charge (b < r) case. In summary, the electric field inside the spherical shell (r < a) is 0, between the inner and outer radii (a < r < b) it is given by E = (4πkρ(b^3 - a^3))/3r^3, and outside the outer radius (r > b) it is given by E = kQ/r^2.
Liquidxlax

## Homework Statement

A spherical shell of inner radius a and outer radius b contains a uniform charge density ρ of total charge Q. (a) Find ρ (b) find the electric field for r< a, a< r< b, b< r sketch it

E = kQ/r2

V=4πr3

## The Attempt at a Solution

ρ π

since it is a thick shell the volume would be V = (4/3)π(b3 - a3)

and ρV = Qenc => ρ=Q/V

so

ρ = 3Q/(4π(b3 - a3))

(b) r<a the electric field will be 0 because the electric field can't close on itself

a< r< b

E=kQenc/r2 = (4πkρ(b3 - a3))/3r3

b<r kQ/r2 because the sphere can be treated as a point charge and the electric field is symmetric on a sphere

Does this look right?

Liquidxlax said:
E= [...] = (4πkρ(b3 - a3))/3r3
There's two errors with the above equation (see the red).

collinsmark said:
There's two errors with the above equation (see the red).

okay i see how the r3 is wrong, that was just a transcribing error, but how come the b is wrong? or would it be because we know that r<b so we can assume it to be

(r3-a3) because a is constant and r does not exceed b

There you go! Remember Gauss' law relates to the charge enclosed within the Gaussian surface (the Gaussian surface in this case has radius r). In the case where b > r > a, then any charge between r and b is outside the Gaussian surface, and has no effect (in this spherically symmetrical case).

Your attempt at a solution looks mostly correct. However, there are a few things that could be clarified or expanded upon:

1. In part (a), it would be helpful to explicitly state that Qenc = Q, since the shell contains all of the charge Q.

2. In part (b), you could also mention that the electric field inside the shell (a < r < b) is 0 because the charges within the shell cancel each other out, resulting in no net electric field.

3. For the electric field outside the shell (r > b), it would be more accurate to say that the electric field is the same as that of a point charge with charge Q located at the center of the sphere. This is because the charges within the shell can be treated as a point charge at the center of the shell, and the electric field due to this point charge is the same as the electric field outside the shell.

Overall, your solution demonstrates a good understanding of the problem and the relevant equations. Well done!

## What is a thick spherical shell confirmation of solution?

A thick spherical shell confirmation of solution is a method used in physics to confirm the validity of solutions to problems involving thick spherical shells. It involves checking that the solution satisfies the boundary conditions and equations of motion for a thick spherical shell.

## Why is a thick spherical shell confirmation of solution important?

A thick spherical shell confirmation of solution is important because it ensures that the solution to a problem is accurate and valid. This is crucial in physics, where small errors can have significant consequences.

## What are the steps involved in a thick spherical shell confirmation of solution?

The steps involved in a thick spherical shell confirmation of solution include setting up the problem and identifying the relevant equations, applying boundary conditions, solving for the unknown variables, and checking that the solution satisfies the equations of motion and boundary conditions.

## What types of problems can be solved using a thick spherical shell confirmation of solution?

A thick spherical shell confirmation of solution can be used to solve problems involving thick spherical shells, such as finding the electric field or potential inside and outside a charged spherical shell, or determining the stress and strain on a thick spherical shell under external forces.

## Are there any limitations to using a thick spherical shell confirmation of solution?

One limitation of using a thick spherical shell confirmation of solution is that it assumes the shell is perfectly spherical and homogeneous. In reality, real-world shells may deviate from these assumptions, which can affect the accuracy of the solution. Additionally, this method may not be applicable to more complex problems involving non-uniform shells or multiple shells.

• Introductory Physics Homework Help
Replies
2
Views
1K
• Introductory Physics Homework Help
Replies
7
Views
3K
• Introductory Physics Homework Help
Replies
3
Views
757
• Introductory Physics Homework Help
Replies
44
Views
2K
• Introductory Physics Homework Help
Replies
5
Views
4K
• Introductory Physics Homework Help
Replies
7
Views
2K
• Introductory Physics Homework Help
Replies
12
Views
3K
• Introductory Physics Homework Help
Replies
10
Views
2K
• Introductory Physics Homework Help
Replies
16
Views
3K
• Introductory Physics Homework Help
Replies
6
Views
2K