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Electric Field Intensity

  1. Oct 29, 2015 #1
    1. The problem statement, all variables and given/known data
    Show that the electric field E outside a spherical shell of uniform charge density ρs is the same as due to the total charge on the shell located at the centre.

    2. Relevant equations
    Using only Coulomb's Law
    E=Q/4πε0 ar

    3. The attempt at a solution
    If i assumed it as circular disc, ρ will increase indefinitely as radius decreases.
    Many examples shown proof using Gauss's Law but this is a question from first chapter on Coulomb Forces and Electric Field Intensity.
  2. jcsd
  3. Oct 30, 2015 #2

    rude man

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    Very difficult to do if you can't use Gauss's law! Hope you're good at integration ...
  4. Oct 30, 2015 #3
    Thanks. I guess it won't be so difficult since it is question no# 10. I've done the rest till no #20.
    From notes:
    The force field in the region of isolated of charge Q is spherically symmetric.

    Is by proving the E at the surface equal to E if Q is placed at the center, then we prove the given statement?
  5. Oct 30, 2015 #4

    rude man

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    Yes. If you can determine E at the surface without Gauss then I agree that constitutes proof.
  6. Nov 4, 2015 #5
    Finally i got it. Using Mathematica for the integral part.

    dE= dQ/4πε0 aR/R2
    dQ=ρr2Sinθ dθ dΦ
    R2=z2+r2-2rzCosθ - Using Cosine Law
    Zaz=(z-rCosθ)az - both x and y axis cancel.

    E=∫ ∫ ρr2Sinθ dθ dΦ (z-rCos) /(4πε0 az/R3/2
    θ - 0 to π
    Φ-0 to 2π

    Using Mathematica the integral , I
    I=-(r-z)/z2 Ir-zI + (r+z)/z2Ir+zI
    If z<r, I=0 then E=0 , inside the shell.
    If z>r, I=2/z2 then E=ρ4πr2/4πε0z2=Q/4πε0 az/z2
    Last edited: Nov 4, 2015
  7. Nov 7, 2015 #6

    rude man

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    This is very impressive! Fine work.
    I was wondering if it would have been a bit easier to compute the potential, then E = -∇V. Might try it myself. But your work looks fine, congrats!
  8. Nov 7, 2015 #7
    Thank you. Just started reading electromagnetic with Coulombs law.
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