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!
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$\begingroup$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".
$\endgroup$ 3 $\begingroup$We assume that the sphere of radius $R$ centered at 0. Let us assume an observation point $o$ above the north pole of the sphere (by symmetry this should provide a good answer). We consider rings from the bottom up to the north polo $(0,0,R)$. The ring at a high $z$, $-R \le z \le R$ has a radius $\rho=\sqrt{R^2 - z^2}$. We prefer to see the problem as a function of the polar angle from $-\pi/2$ to $\pi/2$. We have that $\rho=R \sin \theta$ with $\theta$ the polar angle. It is well known that for a ring with uniform charge density $\sigma$ , radius $r$ and an observation point in the axis of the ring at a distance $d$ from the center (in the direction of the axis of the ring) produces the field
\begin{eqnarray*} E(d,\rho)= \frac{\sigma \rho \, d}{2 \epsilon_0(\rho^2 + d^2)^{3/2}}. \end{eqnarray*} The distance between the observation point $o$ and the ring at $z$ height is $d=o-z$, and $z=R \sin \theta$, then we find
\begin{eqnarray*} E(\theta)= \frac{\sigma (o-R \sin \theta) R \sin \theta}{2 \epsilon_0(R^2 \sin^2 \theta + (o-R \sin \theta)^2)^{3/2}}. \end{eqnarray*} We need to integrate along $\theta$ between $0$ and $\pi$. Along the polar axis the element of integration is $d \ell = R \, d \theta$, so we will need to multiply by $R \, d \theta$.
\begin{eqnarray*} E = \frac{\sigma}{2 \epsilon_0} \int_{-\pi/2}^{\pi/2} \frac{(o-R \sin \theta) R^2 \sin \theta}{(R^2 \sin^2 \theta + (o-R \sin \theta)^2)^{3/2}} d \theta. \end{eqnarray*} Let us perform the following substitution
\begin{eqnarray*} u= \cos \theta \quad , \quad du=- \sin \theta d \theta \\ \theta=0 \implies u = 1\\ \theta=\pi \implies u = -1 , \end{eqnarray*} then
\begin{eqnarray*} E = \frac{\sigma R^2}{2 \epsilon_0} \int_{-1}^1 \frac{o - u R}{(R^2 + o^2 - 2 o R u)^{3/2}} du \end{eqnarray*} We split the integrand in two fractions (forget the coefficient for now).
\begin{eqnarray*} \int_{-1}^1 \frac{o} {(R^2 + o^2 - 2 o R u)^{3/2}} du \quad \mathrm{and} \quad -\int_{-1}^1 \frac{u R} {(R^2 + o^2 - 2 o R u)^{3/2}} du \end{eqnarray*} For the first integral, let us make $x=R^2 + o^2 - 2 o R u$, then $dx=-2 o R du$, and in terms of $x$,
\begin{eqnarray*} -\frac{1}{2 R} \int \frac{dx}{x^{3/2}} = \frac{1}{ R \sqrt{x} }, \end{eqnarray*} Then the first integral is
\begin{eqnarray*} \int_{-1}^1 \frac{o} {(R^2 + o^2 - 2 o R u)^{3/2}} du &=& \left . \frac{1}{R \sqrt{R^2 + o^2 - 2 o R u}} \right |_0^1 \\ \end{eqnarray*}
Let us do the second integral usig integration by parts. We write
\begin{eqnarray*} -\int_{-1}^1 \frac{u R} {(R^2 + o^2 - 2 o R u)^{3/2}} du &=& -\frac{u }{o \sqrt{R^2 + o^2 - 2 o R u}} \\ && + \int \frac{1}{o \sqrt{R^2 + o^2 - 2 o R u}} du \end{eqnarray*} Now,
\begin{eqnarray*} \int \frac{1}{o \sqrt{R^2 + o^2 - 2 o R u}} du = -\frac{1}{o^2 R} \sqrt{R^2 + o^2 - 2 o R u}, \end{eqnarray*} then
\begin{eqnarray*} \int_{-1}^1 \frac{u R} {(R^2 + o^2 - 2 o R u)^{3/2}} du &=& \frac{u }{o \sqrt{R^2 + o^2 - 2 o R u}} + \frac{ \sqrt{R^2 + o^2 - 2 o R u}}{o^2 R} \\ &=& \frac{ R^2 + o^2 - o R u}{o^2 R \sqrt{R^2 + o^2 - 2 o R u}} \end{eqnarray*} Putting the first and the second integrals back together we get
\begin{eqnarray*} \frac{1}{R \sqrt{R^2 + o^2 - 2 o R u}} - \frac{ R^2 + o^2 - o R u}{o^2 R \sqrt{R^2 + o^2 - 2 o R u}} = \frac{-R^2 + o Ru}{o^2 R \sqrt{R^2 + o^2 - 2 o R u}} \end{eqnarray*} Hence we found that
\begin{eqnarray*} \int \frac{o - u R}{(R^2 + o^2 - 2 o R u)^{3/2}} du = \frac{o u - R}{o^2 \sqrt{o^2 - 2 o R u + R^2}}, \end{eqnarray*} and so
\begin{eqnarray*} \left . \frac{o u - R}{o^2 \sqrt{o^2 - 2 o R u + R^2}} \right |_{-1}^1 &=& \frac{o - R}{o^2 \sqrt{o^2 - 2 o R + R^2}} + \frac{o + R}{o^2 \sqrt{o^2 + 2 o R + R^2}} \\ &=& \frac{o - R}{o^2 |o - R|} + \frac{o + R}{o^2 |o + R|} \end{eqnarray*}
\begin{eqnarray*} E= \frac{\sigma R^2}{2 \epsilon_0} \left [ \frac{o - R}{o^2 |o - R|} + \frac{o + R}{o^2 |o + R|} \right ]. \end{eqnarray*} That is
\begin{eqnarray*} E = \left \{ \begin{array}{cc} \frac{\sigma R^2}{o^2 \epsilon_0} & o > R \\ \\ 0 & o < R \end{array} \right . \end{eqnarray*} but
\begin{eqnarray*} \frac{\sigma R^2}{o^2 \epsilon_0} = \frac{4 \pi \sigma R^2}{4 \pi o^2 \epsilon_0} = \frac{Q}{4 \pi o^2 \epsilon_0} \end{eqnarray*} where $4 \pi R^2 \sigma$ is the total charge in the sphere. Then
\begin{eqnarray*} E = \left \{ \begin{array}{cc} \frac{Q}{4 \pi o^2 \epsilon_0} & o > R \\ \\ 0 & o < R \end{array} \right . \end{eqnarray*}
What if $o=R$?
$\endgroup$ $\begingroup$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.
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