Let the electrical field inside the spherical shell be $E_1$ and outside be $E_2$, then by Gauss law, $E_2 – E_1 = 4\pi \sigma /k$, where $\sigma$ is surface charge density of the spherical shell.
Let the the volume charge density be $\rho$ ranging from $x = 0$ to $x = x_0$ the thickness of the shell. Consider a much thinner slab of thickness $dx << x_0$.
Then, the force is given by $$F = \int_0^{x_0} E \ \rho \ dx$$
The small change in $E$ is given by $dE = 4\pi\rho dx/k$,
Thus we get $$F = {k\over 4\pi} \int_{E_1}^{E_2} E \ dE = {k\over 8\pi}(E_2^2 – E_1^2) = …$$
- In this proof I did not understand how they got $E_2 – E_1 = 4\pi\sigma/k$, I know that is true for a spherical shell with $0$ thickness, but how did they prove it in general ?
- If $\rho$ is volume charge density then its dimensions are $[CL^{-3}]$, where $C$ is dimension for charge, then multiplying by length $(dx)$ won't give the dimensions of charge. So how did they say that charge is $\rho dx$ ?
Best Answer
According to the resource you linked (Electricity and Magnetism, Purcell - https://i.stack.imgur.com/rZxJq.jpg), considering a sufficiently small patch of the surface that can be approximated to be flat, which has surface A and not negligible thickness, you can still define surface density as $$\sigma = \frac{Q}{A}=\frac{\int\rho\,dV}A=\int\rho\,\frac{dV}A=\int\rho\,dr.$$ Here Q is the total charge in the patch (the prism that has base A and height $\Delta r$, referring to your resource).
Gauss's law applied to a sufficiently small patch of area A says in general that $$E_2\,A-E_1\,A=\frac{4\pi}{k}Q$$ [see https://drive.google.com/file/d/0BxaPElogYL4cSmdMc0JGQWthQk0/view, but in this case you do not need to consider the limit as $\epsilon \to 0$ because you are considering a patch that has a really small area, so that is almost flat, and so that the field it generates is almost like the one of an infinite flat thick surface charge; this, as you know or as you may want to verify, generates a field perpendicular to the surface both inside and outside the surface, so when you consider the "Gaussian pillbox" the field is parallel to the sides, and the only contribution to flux is from the top and the bottom].
Applying to this particular case, $$\Delta E=\frac{4\pi}{k}\frac{Q}{A}=\frac{4\pi\sigma}{k}.$$
About point 2, you are calculating force per unit area, that is electrostatic pressure.