# Homework Help: Find electric field strength expression for a specfic region

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1. Feb 8, 2017

### jlmccart03

1. The problem statement, all variables and given/known data
A thick, spherical shell of inner radius a and outer radius b carries a uniform volume charge density ρ.

Find an expression for the electric field strength in the region a<r<b.

2. Relevant equations
Gauss's Law ∫E⋅dA
Area of a sphere

3. The attempt at a solution
I know I am really close to the answer, but I do not know what I am doing wrong. I know that there are two radii, but how do I find the electric field strength with these. I think I am close, but may be way off too. Can someone guide me to the way of solving this.

2. Feb 8, 2017

### Staff: Mentor

What has been your approach so far? Can you show us an attempt?

3. Feb 8, 2017

### jlmccart03

Ok this is my attempt: (1/4πr^2)*(Qenclosed/ε0) = (1/4πr^2)*(πa^2/ε0) = (1/4r^2)(a^2/ε0). Correct?

4. Feb 8, 2017

### Staff: Mentor

How did you determine Qenclosed? I don't see the charge density ρ used, and a measure of volume should involve the cube of a radius (or two). Perhaps you should show in detail how you determined Qenclosed.

5. Feb 8, 2017

### jlmccart03

Well, Qenclosed is simply πa^2ρ. I forgot to put in the ρ in the previous post. Its this because of the sphere. I just used a previous example where in class we used a cylinder instead and got it to be πr^2L for Qenclosed without the ρ. I may be confused though on this part.

6. Feb 8, 2017

### Staff: Mentor

The enclosed charge should go as the volume of the portion of the thick shell, so expect to see terms that are radius cubed, not squared. πa^2ρ having only a radius squared term indicates that it's an area element, not a volume. I think you need to think about how you might determine the volume of a portion of the shell.

7. Feb 8, 2017

### jlmccart03

Wait a minute. I thought I am dealing with area and not volume. Since I am dealing with Gauss's law which is ∫E⋅dA. I didn't think volume was a part of this question at all. Am I wrong?

8. Feb 8, 2017

### Staff: Mentor

You need the surface area of the Gaussian surface, and you also need the enclosed charge. The enclosed charge is spread over a volume, a portion of the thick shell. It's the subshell extending from radius $a$ to radius $r$.

9. Feb 8, 2017

### jlmccart03

Ok so in that case how do I apply the volume of a sphere being 4/3πr^3? Does the 4's cancel leaving (1/3r^3)(a^2/ε0ρ).

10. Feb 8, 2017

### Staff: Mentor

A shell is not a solid sphere, so its volume won't be that of a solid sphere.

First work out how to determine the volume of a spherical shell (the shell itself, not including the empty cavity that the shell encloses). The result will include variables a and r, with a being the inside surface radius of the shell and r some radius between a and b.

11. Feb 8, 2017

### jlmccart03

Ok so I tried my own way of getting the answer and came up with ((b2-a2)/4r3ε0))ρ.

12. Feb 8, 2017

### Staff: Mentor

It is incorrect. The units do not resolve to those of an electric field (N/C). They resolve to something like N/(C*m^2) instead.

I think you're still getting tripped up by the volume calculation. I would expect the result to involve cubes in the numerator and a square in the denominator (corresponding to volume in the numerator as part of the calculation of the enclosed charge, and area in the denominator for the area of the Gaussian surface).

Have you tried doing a web search to find the volume of a spherical shell so that you can see what you're trying to match?

13. Feb 8, 2017

### jlmccart03

So its V = (4/3)π(r3-(r-t)3). I need it in this form?

14. Feb 8, 2017

### Staff: Mentor

Something like that, although I don't know what t is. It's not one of the variables defined in the problem.

I'll give you a hint: If you take the volume of a solid sphere and take away the volume of a slightly smaller sphere inside it (leaving an empty spherical cavity in its place), what sort of object do you end up with?

In this case the slightly smaller sphere represents the hollow cavity inside your shell (so radius a). The larger sphere represents the solid sphere out to the Gaussian surface that you've drawn within the original shell (radius r).

Last edited: Feb 8, 2017
15. Feb 8, 2017

### jlmccart03

So its something like (4/3)π(b3-(b-x)3)-(4/3)π(a3-(a-x3)? Its similar to a washer problem in Calculus correct?

16. Feb 8, 2017

### Staff: Mentor

It is similar, yes. But you don't need to introduce new variables. You already have the two radii, a and r.

17. Feb 8, 2017

### jlmccart03

But b is the outer radius and a is the inner radius, I actually have no clue where r is placed.

18. Feb 8, 2017

### Staff: Mentor

From the problem statement:

19. Feb 8, 2017

### jlmccart03

Sorry I didn't write it down, but the problem states to "Express your answer in terms of r, a, b, ρ, ϵ0." It was in the picture.

Also it says "A thick, spherical shell of inner radius a and outer radius b carries a uniform volume charge density ρ."

20. Feb 8, 2017

### Staff: Mentor

Yes, and r is in between a and b. a and r are the only two radii that are important (so long as r never exceeds b).

21. Feb 8, 2017

### jlmccart03

OHHHHH ok, so I just use a the inner radius and r the radius for the thick piece to find the expression. So once I subtract the two volumes I get the volume of the inner piece. From there do I do the basic Qenclosed/ε0 or is there another additional step?

22. Feb 8, 2017

### Staff: Mentor

No that's it. Just apply Gauss' Law to the enclosed charge.

23. Feb 8, 2017

### jlmccart03

So I should get (((r-a)x-r^2+b^2))x/ε0r^2)ρ. Correct?

24. Feb 8, 2017

### Staff: Mentor

No. there will be cubes in the numerator; You're dealing with volumes. And there will be no variable x. There will be variables a, r, ρ, and ε0 in the answer.

How are you calculating the volume of the shell of interest? Can you show your derivation?

25. Feb 8, 2017

### jlmccart03

The volume should actually be (4/3)π(b3-(b-r)3). Then I took that and applied (1/4πε0) to get r(r2-3br+3b2)/(3ε0)