Integrating the spherical shell

This page describes how to integrate a uniformly charges spherical shell in order to prove that it will look like a point charge from the outside but will have a zero electric field on the inside. The understanding of this is important because there are many objects that have a charge on the outside but have zero electric field on the inside.

The Main Idea

The electric field of a conducting sphere can also be found using Gauss' Law. Using Gauss's Law, you model a Gaussian surface of a sphere with radius r>R and the electric field will have the same magnitude, directed outward, at every point on the surface of the sphere.

This page serves more as a proof on understanding why there is an electric field outside the sphere, but not inside.

A Mathematical Model

Assume the spherical shell is centered at the origin.

Step 1: divide the sphere into (ring like) pieces. Imagine the sphere is sliced into several different rings. You must have these rings measure some distance 'theta' from the middle.

for example: [math]\displaystyle{ {ϴ, ϴ + Δϴ} }[/math] ... etc

Each ring contributes delta E at an obseration point some distance r greater than the radius R away from the center of the sphere.

Step 2: compute the distance of the ring from the observation location. (remember its :observation location - source) [math]\displaystyle{ {d = (0-Rcos(ϴ)} }[/math] find the amount of charge on each ring: [math]\displaystyle{ {ΔQ}={\frac{surface area of ring}{surface area of sphere}}=Q{\frac{2pi(Rsinϴ)(RΔϴ)}{4piR**2}} }[/math]

Note that the radius of the ring is Rsin(ϴ)' and its width is RΔϴ. The integration variable is: ϴ The Magnitude of ΔE is: [math]\displaystyle{ {ΔE}={\frac{1}{4piε0}}*{\frac{ΔQd}{d**2+((Rsinϴ)**2)**(3/2)}}={ΔE}={\frac{1}{4piε0}}**{\frac{r-Rcosϴ}{(r-(Rcosϴ)**2)+(Rsinϴ**2)**(3/2)}}Q{\frac{2pi(Rsinϴ)}{4piε0}}*{RΔϴ} }[/math]

Step 3: Add the pieces together [math]\displaystyle{ {ΔE}={\frac{1}{4piε0}}*{\frac{Q}{2}}*(integral from 0 to pi of:){\frac{r-Rcosϴ}{(r-(Rcosϴ)**2)+(Rsinϴ**2)**(3/2)}}{sinϴ*Δϴ} }[/math]

I would recommend using wolfram alpha to evaluate the integral as it wan get a little difficult.

The results for outside of the shell are: [math]\displaystyle{ {ΔE}={\frac{1}{4piε0}}*{\frac{Q}{r**2}} }[/math]for r>R for inside the shell: (r<R)[math]\displaystyle{ {E=0} }[/math]

See also

A Spherical Shell of Charge

Further reading

Matter and interactions 4th edition by Chambay and Sherwood

References

Matter & Interactions Volume II