1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Find electric field strength expression for a specfic region

  1. Feb 8, 2017 #1
    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.
    upload_2017-2-8_13-27-29.png

    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. jcsd
  3. Feb 8, 2017 #2

    gneill

    User Avatar

    Staff: Mentor

    What has been your approach so far? Can you show us an attempt?
     
  4. Feb 8, 2017 #3
    Ok this is my attempt: (1/4πr^2)*(Qenclosed/ε0) = (1/4πr^2)*(πa^2/ε0) = (1/4r^2)(a^2/ε0). Correct?
     
  5. Feb 8, 2017 #4

    gneill

    User Avatar

    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.
     
  6. Feb 8, 2017 #5
    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.
     
  7. Feb 8, 2017 #6

    gneill

    User Avatar

    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.
     
  8. Feb 8, 2017 #7
    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?
     
  9. Feb 8, 2017 #8

    gneill

    User Avatar

    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##.
     
  10. Feb 8, 2017 #9
    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).
     
  11. Feb 8, 2017 #10

    gneill

    User Avatar

    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.
     
  12. Feb 8, 2017 #11
    Ok so I tried my own way of getting the answer and came up with ((b2-a2)/4r3ε0))ρ.
     
  13. Feb 8, 2017 #12

    gneill

    User Avatar

    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?
     
  14. Feb 8, 2017 #13
    So its V = (4/3)π(r3-(r-t)3). I need it in this form?
     
  15. Feb 8, 2017 #14

    gneill

    User Avatar

    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
  16. Feb 8, 2017 #15
    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?
     
  17. Feb 8, 2017 #16

    gneill

    User Avatar

    Staff: Mentor

    It is similar, yes. But you don't need to introduce new variables. You already have the two radii, a and r.
     
  18. Feb 8, 2017 #17
    But b is the outer radius and a is the inner radius, I actually have no clue where r is placed.
     
  19. Feb 8, 2017 #18

    gneill

    User Avatar

    Staff: Mentor

    From the problem statement:

     
  20. Feb 8, 2017 #19
    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."
     
  21. Feb 8, 2017 #20

    gneill

    User Avatar

    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).
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted