Like James Maslek said, this is just an effect of having an infinitely thin disk--the field is a step function.
If you like, you can replace your disk with two disks , each having the same radius and surface charge density $\sigma/2$. Seperate them by some finite distance, so they are located at $z = \pm \epsilon$ for some small $\epsilon$. Then, work out what the field is at some arbitrary point along their common axis. You will find that the field at $z=0$ is zero for all $\epsilon$ and that your solution limits to your original equation for $z \gg \epsilon$. This way, you can ignore the discontinuous nastiness of the infinitely thin disk.
If you want to get yet fancier, and REALLY analyze this situation in the case where the thickness of the disk is important, you can think of a cylinder with uniform volume charge density, radius $r$ and thickness $L$. You can work out the field at an aribtrary point along the cylinder's axis, inside or outside. It should be easy enough to show that the field at the center of the cylinder is zero. Then, you can replace $\rho$ with $\frac{\sigma}{L}$ and see how it compares with your expression for a disk in the limit $z \gg L$, which should once again equal your original expression.
Finally, a third way to see this, but one which involves much harder calculus would be to derive the field due to a disk at an arbitrary point on the plane of the disk, and then see that the field goes to zero at the center of the disk when you are constrained to the plane.
- How can really a field exert force on the charges that created the field?
I think the point of the passage is to assume that they do not exert forces on themselves. They want to compute the force on the little patch due to everything else, they assume the patch exerts no net force on itself and therefore compute the force due to the whole spherical shell (patch included) and hope this is equal to the force on the patch due to every other part of the spherical shell.
- What does Mr.Purcell mean when saying"the force on due to all other charges within the patch itself.
If you assume the forces the charges feel are due to all the charges then they are the forces due to all charges not on the patch plus the forces due to the charges in the patch. $F_\text{all}=F_\text{from patch}+F_\text{not from patch}.$ If $F_\text{from patch}$ is zero and $F_\text{all}$ is easy to compute then you can use those two (alleged) acts to compute $ F_\text{not from patch}.$
- Why is the electric field average of that of inside & outside?
Imagine you had a thin shell of finite thickness $t$ with a uniform charge density in the shell. Then the field outside is still $Q/4\pi\epsilon_0r^2,$ radially outwards where $r$ is the distance from the center and inside the field is still zero. Both results follow from spherical symmetry and Gauss' Law. But within the shell the field continuous goes from one to the other. For a distance $s$ from the center, spherical symmetry and Gauss' Law again gives, $E(s)4\pi s^2=\oint\vec E \cdot d\vec A=Q_\text{enc}/\epsilon_0=\int_{R-t}^R\rho 4\pi r^2dr/\epsilon_0,$ where $\rho=Q/\int_{R-t}^R\rho 4\pi r^2dr.$
Thus $\rho=\frac{3Q}{4\pi(R^3-(R-t)^3)}$ is a constant and $E(r)=\rho\frac{r^3-(R-t)^3}{3\epsilon_0r^2},$ where $r$ is the distance from the center when you are in the shell. You could then compute the force on each layer of the patch by integrating $\int_{R-t}^RdF=\int_{R-t}^R E(r)dq=dA\int_{R-t}^R E(r)\rho dr.$ So, the force equals $dA\rho\int_{R-t}^R E(r)dr=$
$$dA\rho\int_{R-t}^R \rho\frac{r^3-(R-t)^3}{3\epsilon_0r^2}dr=$$
$$\frac{ dA\rho^2}{3\epsilon_0}\int_{R-t}^Rr-(R-t)^3r^{-2}dr=$$
$$ \frac{dA\rho^2}{3\epsilon_0}\left(
\frac{R^2-(R-t)^2}{2}
+(R-t)^3 \left(\frac{1}{R}-\frac{1}{R-t}\right)\right)$$
$$ =\frac{dA\rho^2}{3\epsilon_0}\left(
\frac{2Rt-t^2}{2}
+(R-t)^3\frac{R-t-R}{R(R-t)}\right)=$$
$$ \frac{dA\rho^2}{3\epsilon_0}\left(
\frac{2Rt-t^2}{2}
-(R-t)^2\frac{t}{R}\right)=$$
$$ \frac{dA\rho^2}{3\epsilon_0}\left(
\frac{2Rt-t^2}{2}
-\frac{(R^2-2Rt+t^2)t}{R}\right)=$$
$$ \frac{dA\rho^2}{3\epsilon_0}\left(
\frac{-t^2}{2}
-\frac{(-2Rt+t^2)t}{R}\right)=$$
$$ \frac{dA\rho^2t^2}{\epsilon_0}\left(
\frac{1}{2}
-\frac{t}{3R}\right).$$
This is the actual force on the shell of finite thickness t, if $t< <R$ then the second term is dominated by the 1/2 so can be neglected. And for a thin shell $\rho t \approx \sigma$ so we get $F_\text{patch}\approx dA\sigma \frac{\sigma}{2\epsilon_0}=dq\frac{\sigma}{2\epsilon_0}$=$dqE_\text{eff}.$ So we see $E_\text{eff}\approx \frac{\sigma}{2\epsilon_0}$ but that this systematically overestimates the force, but in a way that approaches the correct answer as the thickness get smaller and smaller. For a real shell, you could always make it an atomic thickness so that this approximation is swamped by the approximation of the macroscopic field that smoothly varies from place to place instead of shooting massively towards an electron near every place there is an electron and shooting massively away from a proton near every place where there is a proton.
So there is truly nothing wrong with averaging the two fields ... if the shell is very very very thin.
To avoid any confusion about the force we are trying to compute, it is the total force the patch feel due to other charges, i.e. the sum of the forces each part of the patch feels due to the other patches. Morally you want the total force only to include the force of interaction.
If you want the force on every part of the patch and you think the patch exerts equal and opposite forces on itself then when each part exerts forces on itself then you can compute the total force on every part of the patch due to everything else and it equals the thing you want.
Try a much simpler example. Three charges, two of them (q1, q2) in a patch, and one (q3) outside the patch.
You want to compute the force on the patch due to the non patch. This means you want the force between q1 and q3 and the force between q2 and q3 and you want the sum of those two forces. Why? Because that is what we've been asked to compute.
How can we compute it? If we believe that q1 and q2 exert equal and opposite forces on each other then we can compute the force on q1 due to both q2 and q3 and compute the force on q2 due to both q1 and q3 and then when we add them together the forces that q1 and q2 exert on each other cancel out and we get the sum of the force between q1 and q3 and the force between q2 and q3. Exactly what we wanted. Why would we do that? For some problems, the force due to everything is much easier to compute. This is what your text is trying to argue. How reasonable the assumptions are is up to you to judge.
Later you will find that charges exchange momentum with the field and that it takes time for the momentum to flow from where one charge gave the momentum to where another charge receives it, and in the meantime the momentum of the particles does not stay constant, so the third law only holds as conservation of momentum and only when the field has momentum too.
So appealing to the third law without specifying all the flows of momentum is a bit misleading. And the arguments about the fields and forces are for macroscopic fields with continuous charge distributions, again not really reality. But I wanted to present this as telling you what your textbook intended.
Best Answer
The energy required for squeezing the sphere from initial radius $r_0 \to (r_0-dr)$ is not the energy required to assemble the charges. He merely calculated the change in energy (or the work required) each time you squeezed the sphere by a displacement of $dr$. To find the work required for assembling the charges you'll have to integrate over all space (As in you had a sphere of Radius $\to$ infinity and squeeze it to the desired radius).
He found how much work is required to squeeze a sphere a little bit, and again the entire energy required to assemble the charges is to sum over all this little bit of energy from infinity to the desired radius.
It's some how like recursion; Suppose you know the energy $U$ required to assemble the charges at their current state, and now you want to squeeze them a little bit closer so you must invest extra work: $$U(r_0-dr)=U(dr)+U(r)=2U(dr)+U(r+dr)= \cdots =\text{[integration over all} \qquad U(dr) \qquad \text{from infinity to the current radius]} + U\text{(infinity)} \text{[Which is 0]}$$
If you go backwards, and let the electrons move freely to infinity (If they can do that), you'll gain back energy which is basically equal ,in value, to the work required in assembling them in the first place.
I think you have an old version of the book; The one I have has $\dfrac{\sigma}{\epsilon_0}$ instead of that. By the way, there is a "1" near the sigma, there must be an explanation at the bottom of the page I believe.
Hope this helps!