Homework Help: Electric Field Intensity

1. Oct 29, 2015

azizlwl

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. Oct 30, 2015

rude man

Very difficult to do if you can't use Gauss's law! Hope you're good at integration ...

3. Oct 30, 2015

azizlwl

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?

4. Oct 30, 2015

rude man

Yes. If you can determine E at the surface without Gauss then I agree that constitutes proof.

5. Nov 4, 2015

azizlwl

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
E=ρ2πr2/4πε0∫Sinθ(z-rCosθ)/(r2+z2-2zrcosθ)3/2

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
6. Nov 7, 2015

rude man

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!

7. Nov 7, 2015

azizlwl

Thank you. Just started reading electromagnetic with Coulombs law.