- 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
$\def\l{\left}\def\r{\right}$ You can work in the complex plane. The real component is your $\hat i$-component the imaginary component is your $\hat j$-component. There $\l(\cos(\phi),\sin(\phi)\r)$ is represented as $\exp(i\phi)$ with $i=\sqrt{-1}$.
In this context your sum is $$ \sum_{k=0}^{N-1} e^{i2\pi\frac kN}. $$ if we substitute $a := e^{i\frac{2\pi}N}$ into this formula we obtain $$ \sum_{k=0}^{N-1} a^k $$ which can be transformed into a telescope sum by multiplying with $(1-a)$ $$ (1-a)\sum_{k=0}^{N-1} a^k = \sum_{k=0}^{N-1} a^k - \sum_{k=1}^{N} a^k = 1-a^N $$ Thus, we have $$ \sum_{k=0}^{N-1} a^k = \frac{1-a^N}{1-a} $$ But $1-a^N = (e^{i\frac{2\pi}N})^N = e^{i2\pi} = 1$ and we obtain $$ \sum_{k=0}^{N-1} e^{i2\pi\frac kN} = 0. $$
Your sum is zero in exact arithmetics.