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The maximum magnitude of the electric field of a sphere

  1. Dec 4, 2016 #1
    1. The problem statement, all variables and given/known data

    A non-conducting sphere of radius a carries a non-uniform charge density. The electrostatic field inside the sphere is a distance b from is center and is given by E = (b/a)4E0 (E0 being the maximum magnitude of the field.)

    a. Find an expression for E0 in terms of the total charge Q0 and the radius of the sphere.

    b. Determine the charge density of the sphere as a function of radius.


    2. Relevant equations

    ε0 ∫ E ⋅ dA = ∫ ρdV

    dQ = ρdV


    3. The attempt at a solution

    So for a I did ε0 ∫ E ⋅ dA = ε0E4πa2 since
    E = (b/a)4E0
    ε0E4πa2 =
    ε0(b/a)4E0)4πa2 = 4πE0ε0(b4/a2)

    For the ∫ ρdV side of the problem my professor did it a problem similar in his slides for b > a of uniform charge density the ∫ ρdV = Q0 so i just assume it is the same for a non-uniform charge density. Getting me
    4πE0ε0(b4/a2) = Q0
    thus E0 = (a2Q0)/(4πb4)

    For part b I did dQ0 = ρdV = ρ4πa2da so
    Q0 = ∫0b ρ4πa2da = (4/3)πρb^3 so
    ρ = (3Q0)/(4πb3)
     
  2. jcsd
  3. Dec 5, 2016 #2

    haruspex

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    You do not need to assume this. The field you are given is spherically symmetric, so you know that if you decompose it as nested thin spherical shells each shell is uniformly charged. What do you know about the field outside such a shell? You can go straight from that to the answer.
    I don't follow your thinking here. Q0 is a constant, so dQ0 means nothing (or zero).
     
  4. Dec 5, 2016 #3
    You seem to have some misconceptions, so let's just start at the beginning. Take a look at your equation for E and at what radius it will be at its maximum of E0. You should notice that when a=b, you'll have E=E0. What does this tell you about the charge enclosed by your Gaussian surface when you have maximum E?
     
    Last edited: Dec 5, 2016
  5. Dec 5, 2016 #4

    haruspex

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    It would make more sense to write that as ##Q(b)=\int_0^b\rho\pi r^2.dr##, but
    No, to get that ρ would have to be constant.
    Instead, can you find two things that should equate to ##\frac{dQ}{db}##?
     
  6. Dec 5, 2016 #5
    That is typo I meant that when you integrate dQ is comes out to the total charge of Q which is Q0


    I am confused on why i would need to find dQ/db or what dQ/db is supposed to be.


    That when the sphere is at the maximum of E0 the electrostatic sphere is a distance a from the center of the sphere and the charge would be at its maximum.
     
  7. Dec 5, 2016 #6
    Right, the charge is at its maximum. So while using Gauss' Law, the qenc is simply Q0, and the radius of your Gaussian surface is the radius of the sphere. That gives you everything you need to put E0 in terms of Q0 and a, no integration needed.
     
    Last edited: Dec 6, 2016
  8. Dec 6, 2016 #7
    Relooking at what you wrote (it seems clearer now for some reason), you basically have it, you just need to plug in a for b. Then plug what you get for E0 into E, plug that back into Gauss' Law, and you have the result of the charge density integral. So if you have the result of the integral (which is with respect to b) you can find the charge density by taking the derivative of that result with respect to b.
     
    Last edited: Dec 6, 2016
  9. Dec 6, 2016 #8

    haruspex

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    I wrote Q=Q(b), meaning the total charge inside radius b. See my first line in post #4.
     
  10. Dec 7, 2016 #9
    So what you are sayings is I take my answer for a E0 = (a2Q0)/4πb4ε0 since b=a when E0 = E I replace b for a so E0 = Q0/4πa2ε0 Then i take that answer for a and plug it into E so I get E = (Q0b4)/4πa6ε0 Than i plug that into guass's law so ε0∫E⋅dA = (ε04πa2Q0b4)/4πa6ε0 = (Q0b4)/a4 so then (Q0b4)/a4 = ∫ρdV I then take the derivative of both sides and i get 3Q0b3/a4 = (ρ4πb3)/3 so ρ = (9Q0)/4πa4
     
  11. Dec 7, 2016 #10
    Almost. It looks like you're plugging it back into Gauss' law incorrectly. Since E has no angular variance, when you plug it in you should get ##\mathbf E \int_S dA## which, with the integral of the Gaussian surface being that of a sphere, should come out to ##\frac {Q_0 b^4} {4 \pi a^6 \epsilon_0} 4 \pi b^2 = \frac 1 {\epsilon_0} \int_V \rho dV##.
     
  12. Dec 7, 2016 #11
    Why when i take the surface integral of dA would it come out to 4πb2 instead of 4πa2 isn't the area supposed to be base of the radius of the sphere?
     
  13. Dec 7, 2016 #12
    It's the radius of your Gaussian Surface, which in this case is any arbitrary radius equal to or less than a (since you're only interested in values of the sphere to find the charge density, and your equation for the electric field is only good for those values anyways).
     
  14. Dec 7, 2016 #13
    So would I do (Q0b6)/a6 = ∫ρdV then i take the derivative of both sides so (6Q0b5)/a6 = (ρ4πb3)/3 so
    ρ = (9Q0b3)/2a6
     
  15. Dec 7, 2016 #14
    Yes, but when you take the derivative of the integral, you're basically just canceling it out.
     
  16. Dec 7, 2016 #15
    I just realized I keeping trying to turn ∫dV into V like i am taking the integral. So it would just be
    ρ = (6Q0b5)/a6
     
  17. Dec 7, 2016 #16
    Yep, good job!
     
  18. Dec 7, 2016 #17
    Thanks for the help!
     
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