1. Not finding help here? Sign up for a free 30min 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!

Nonconducting spherical shell with uniform charge

  1. Sep 3, 2016 #1
    1. The problem statement, all variables and given/known data
    Suppose the nonconducting sphere of Example 22-4 has a spherical cavity of radius r1 centered at the sphere's center (see the figure). Assuming the charge Q is distributed uniformly in the "shell" (between r = r1 and r = r0), determine the electric field as a function of r for the following conditions.

    r1 < r < r0

    22-31.gif

    2. Relevant equations
    Gauss's law

    3. The attempt at a solution

    I know the electric field is 0 in the inner cavity. The electric field outside of the shell is Q/(4pi*E0*r^2). How would I derive the equation for the electric field inside the shell? Charge per unit volume should be p=Q/(4/3*pi(r0-r1)^3) right?
     
    Last edited: Sep 4, 2016
  2. jcsd
  3. Sep 3, 2016 #2

    SammyS

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    That charge density is incorrect.

    A3 - B3 ≠ (A - B)3 .
     
  4. Sep 3, 2016 #3
    Mmm yes, your'e right it should be Q/(4pi/3(r0^3-r1^3)). But where do you go from there?
     
    Last edited: Sep 4, 2016
  5. Sep 3, 2016 #4

    SammyS

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    How much charge there interior to a sphere of radius, r, where r1 < r < r0 .
     
  6. Sep 3, 2016 #5
    Could you rephrase that and be more specific?
     
  7. Sep 3, 2016 #6
    Would it be the same as the electric field outside the sphere?
     
  8. Sep 4, 2016 #7

    SammyS

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    No, electric charge is not equal to electric field.

    You want to find what the electric field is interior to the "shell". That's where r1 < r < r0 .

    You mention Gauss's Law.
     
  9. Sep 4, 2016 #8
    Right its

    ∫E⋅da=qenc0

    So I keep getting E4πr^2=qenc0

    so that would mean E would equal the same as the electric field as the outside of the sphere, which I know is wrong. What am I doing wrong?
     
  10. Sep 4, 2016 #9

    SammyS

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    For r between r1 and r0, what do use for your Gaussian surface?
     
  11. Sep 4, 2016 #10
    Would it be r1^2 instead of r^2?
     
  12. Sep 4, 2016 #11

    SammyS

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    No.

    If you want to know the electric field at a distance r from the origin, use s sphere of radius r.

    How much electric charge is enclosed within this sphere?

    In the case of your problem the only charge inside this sphere resides in the spherical shell with inner radius r1 and outer radius r. Right ?

    So, how much charge is that ?
     
    Last edited: Sep 4, 2016
  13. Sep 4, 2016 #12
    So if charge density =

    Q/ (4π/3 ( r03-r13)) = qenc/(4π/3 (r3-r13)).

    Then,

    Q(r3-r13) = qenc (r03-r13)

    qenc = [Q(r3-r13)] / (r03-r13)

    E4πr2 = Q(r3-r13)] / ε0(r03-r13)

    E = Q(r3-r13)] / 4πr2ε0(r03-r13)
     
    Last edited: Sep 4, 2016
  14. Sep 4, 2016 #13
    Can someone verify this?
     
  15. Sep 4, 2016 #14

    SammyS

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    Patience.

    That's the right idea. To be perfectly correct, the entire denominator should be enclosed in parentheses, else use some other grouping symbol.
     
  16. Sep 4, 2016 #15
    Haha, thank you SammyS!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted