Electric field of a spherical shell

Question

I hope this question is appropriate for this site, if not, just leave a comment and I will delete.

I am interested in knowing how to derive the electric field due to a spherical shell by Coulomb's law without using double integrals or Gauss Law.

Relevant equations are -- Coulomb's law for electric field and the volume of a sphere:

$\vec{E} = \frac{1}{4\pi \epsilon_0}\frac{Q}{r^2}\hat{r}$, where $Q =$ charge, $r=$ distance.

$V = \frac{4}{3}\pi r^3.$

From my book, I know that the spherical shell can be considered as a collection of rings piled one above the other but with each pile of rings the radius gets smaller and smaller.

I am not interested in the final formula, just the derivation of it. Thank you!

Answer 1

Well you can first calculate the field of a ring centered at $z=z_0$ on the $z$ axis with radius $r$ (using CGS, multiply by ugly factors later). By symmetry, on the $z$ axis the field is only in the $z$ direction and can be shown to be: $$E_z(z)=\frac{q(z-z_0)}{((z-z_0)^2+r^2)^{3/2}}$$ Now each ring has charge $q=Q\cos \theta d\theta$, and $z_0 = R\cos \theta$. This means you can integrate the expression $E_z(z)$ over $\theta$ to get the field at any point on the $z$ axis. By symmetry, you can choose the ring direction as you wish, so that this expression is true for points not on the $z$ axis as well, with $r$ replacing $z$.

As I mentioned in the comments, since the field of each ring contains an integral, this is really a double integral, even if you decide to call this "two single integrals".

Answer 2

You can derive the electric field without using double integrals explicitely, using Gauss law:

$$ \Phi = \epsilon_0 Q $$

Where $\Phi$ is the flow of the electric field across the Gaussian surface. By symmetry you can choose a sphere of radius $R$ bigger than the radius of the charged sphere and the field will be normal and constant on all the surface, so $\Phi = 4\pi R^2 E$, from here you find

$${\bf E} = \frac{Q}{4\pi \epsilon_0 r^2} \hat{\bf r} $$

If one insists in dividing the shpere in rings I see no way to avoid integration.