Electric Field Outside a Uniformly Charged Spherical Shell

[azizlwl]

Homework Statement

Show that the electric field E outside a spherical shell of uniform charge density ρ_s is the same as due to the total charge on the shell located at the centre.

Homework Equations

Using only Coulomb's Law
E=Q/4πε₀ a_r

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.

 

[rude man]

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

 

[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?

 

[rude man]

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?

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

 

[azizlwl]

Finally i got it. Using Mathematica for the integral part.

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

E=∫ ∫ ρr²Sinθ dθ dΦ (z-rCos) /(4πε₀ a_z/R^3/2
θ - 0 to π
Φ-0 to 2π
E=ρ2πr²/4πε₀∫Sinθ(z-rCosθ)/(r²+z²-2zrcosθ)^3/2 dθ

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 a_z/z²

 

[rude man]

Finally i got it. Using Mathematica for the integral part.

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

E=∫ ∫ ρr²Sinθ dθ dΦ (z-rCos) /(4πε₀ a_z/R^3/2
θ - 0 to π
Φ-0 to 2π
E=ρ2πr²/4πε₀∫Sinθ(z-rCosθ)/(r²+z²-2zrcosθ)^3/2 dθ

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 a_z/z²

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!

 

[azizlwl]

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!

Thank you. Just started reading electromagnetic with Coulombs law.