As I tried to answer the SE question Electric fields in continuous charge distribution, I faced many ambiguities regarding this matter. In classical electrodynamics, it is claimed that the electric field near a vast plate is constant (uniform) considering that the charges are continuously distributed (uniformly) all over an infinite plane. The electric field is thus calculated to be:
$$E=\frac{\sigma}{2\epsilon_0} \space ,$$
where $\sigma$ is the surface charge density. However, in reality, the charge is quantized, and the electric field very close to the radius of these particles (a proton for instance) is near to the order of $E\approx 10^{22} \space V/m$ which is extremely high. When I considered $\frac{L}{d}×\frac{L}{d}=\frac{L^2}{d^2}$ number of these particles building an $L×L$ mesh of charges, so that the distance between every two successive charges is $d$ along both $x$ and $y$ axes, I deduced:
$$E=\frac{e}{\pi\epsilon_0} \sum_{n=0}^{\frac{L/2}{d}}\sum_{m=0}^{\frac{L/2}{d}}\space \frac{z}{(n^2d^2+m^2d^2+z^2)^{3/2}} \space$$
where $e$ is the charge of each particle (proton). Moreover, I assumed $L\gg d \gg z$. The above equation shows the field of the mesh, a distance $z$ exactly above the charge located at the center of the plate. However, when $z$ approaches zero, I numerically calculated that $E$ tends to infinity. This obviously violates the classical result of $E=\sigma/(2\epsilon_0)$, which indicates a constant field for a large uniformly charged plane. Does my calculation show that the electric field very close to, say, a plate capacitor is no longer constant but rather extremely great?
If this deduction is correct, the field component of a single particle (proton) parallel to the plate can also get high values very close to each proton, whereas we know that the E-field is zero inside the plate, otherwise we would have current. Where is the problem?
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