Why is electric field of an infinite plate constant at all points? I know from Gauss law, it is $\vec{E}=\dfrac{\sigma}{2 \epsilon_0}(\hat{n})$ at all points. But it doesn't make sense because of the inverse square nature of electric field which suggests if you move further away from the plane, electric field must reduce. 
Then why is electric field of an infinite plate constant at all points?
 A: Here's a quick way to think of it.
Imagine you are distance R from the plate, and you know the force from a circle on the plate that has radius R. The area of the circle is pi R^2.
Now move twice as far. The force from each point charge is reduced from 1/R^2 to 1/4R^2 by the inverse square law. The area of a circle that has radius 2R is 4 pi R^2.
The geometry means the total force stays the same. Every change due to the inverse square law is balanced out by the same change due to the increased area of the homologous structure.
A: The field gets weaker the further you get from a point charge because the field lines can spread out.
The field lines of an infinite plane can never spread out; they just run parallel to each other forever. So the field strength is constant.
A: I think the best way to answer this question is to actually do the math and physics. From first principles and not some shortcut.
From Couloub's law and the definition of the electric field:
$$\vec{E} = \frac{1}{4\pi\epsilon_0} \frac{q}{r^2} \hat{\mathbf{r}}$$
Consider first an infinite wire of change (we will build the sheet later). For now, we assign a charge density of the entire wire: $\lambda$. Where $\lambda = \frac{dq}{d\ell}$.
 
The differential form of the electric field equation may then be given as (using the notation from the image):
$$d\vec{E} = \frac{1}{4\pi\epsilon_0} \frac{dq}{r^2} \hat{\mathbf{r}} = \frac{1}{4\pi\epsilon_0} \frac{\lambda}{r^2} d\ell \;\hat{\mathbf{r}}$$
Now, from the image, it should be a bit clear that the electrical field components from the wire in the "up down" ($\hat{\mathbf{y}}$) direction cancel each other out regardless of the value of $R$ and $\ell$. Therefore, let us only consider the electric field in the $\hat{\mathbf{x}}$ direction.
From the geometry, we notice the following:
$$ r = \sqrt{\ell^2 + R^2} = \frac{R}{\cos \theta} $$
$$ d\ell = R d\theta $$
$$ \hat{\mathbf{x}} = \cos \theta \; \hat{\mathbf{r}}$$
Therefore:
$$d\vec{E} = \frac{1}{4\pi\epsilon_0} \frac{\lambda \cos^2 \theta}{R^2} \frac{R}{\cos \theta} d\theta \;\hat{\mathbf{x}}$$
$$ = \frac{1}{4\pi\epsilon_0} \frac{\lambda}{R} \frac{\cos^2 \theta}{\cos \theta} \;\hat{\mathbf{x}} $$
Now, we want to find the total electric field from the entire length of the wire. Thus, we want to integrate over the entire wire. This means that, intergrating over the angle of $\theta$: $-\frac{\pi}{2} \rightarrow \theta \rightarrow \frac{\pi}{2}$. Therefore:
$$ \vec{E_x} = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{1}{4\pi\epsilon_0} \frac{\lambda}{R} \frac{\cos^2 \theta}{\cos \theta} \;\hat{\mathbf{x}} $$
$$ = \frac{1}{4\pi\epsilon_0} \frac{\lambda}{R} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos \theta \;\hat{\mathbf{x}} $$
$$ = \frac{1}{4\pi\epsilon_0} \frac{2\lambda}{R} \;\hat{\mathbf{x}}$$
Note that, for an infinite wire, the electric field does depend on your distance from the wire. 
However, we want the sheet. We can construct a sheet of chrage by aligning many wires in a row, parallel to each other. Suppose, still using the image, we stack them along the $\hat{\mathbf{z}}$ axis. It should be clear that, like the $\hat{\mathbf{y}}$ component of the electric field cancels itself out when the wire runs along that axis, the sheet also cancels out the contributions from $\hat{\mathbf{z}}$.
We reassign the distance that the point in question is from the sheet as $D$, as $R$ is now between the point and one of the wires (a distance $z$ from the point on the sheet above the point in question) in the entire sheet. This means that $R$ is related now, given by:
$$ R = \sqrt{D^2 + z^2} = \frac{D}{\cos \phi} $$
Where $\phi$ is the angle between the lines $R$ and $D$, similar to how $\theta$ is the angle for the image about (just extrapolate to 3D). Moreover, the surface charge of the sheet is now given by:
$$ \lambda = \sigma dz = \sigma D d\phi $$
$$ \hat{\mathbf{r'}} = \cos \phi \; \hat{\mathbf{x}} $$
Note that the second equation might not make a lot of sense at first; however it is similar to our previous transformation ($ \hat{\mathbf{x}} = \cos \theta \; \hat{\mathbf{r}}$) execpt that the direction is a new offset from $\hat{\mathbf{r'}}$.
If we take the answer for the electric field via a line of charge and put it into a differential form:
$$ d\vec{E_{r'}} = \frac{1}{4\pi\epsilon_0} \frac{2\lambda}{R} \;\hat{\mathbf{r'}} $$
Subsituting:
$$ d\vec{E_x} = \frac{1}{4\pi\epsilon_0} \frac{2 \sigma D }{D} \frac{\cos \phi}{\cos \phi} d\phi \;\hat{\mathbf{x}} $$
$$ = \frac{1}{4\pi\epsilon_0} \left( 2 \sigma \right) d\phi \;\hat{\mathbf{x}} $$
Finally, again, as with the wire, we integrate over the entire sheet: $-\frac{\pi}{2} \rightarrow \phi \rightarrow \frac{\pi}{2}$
$$ \vec{E_x} = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{1}{4\pi\epsilon_0} \left( 2 \sigma \right) d\phi \;\hat{\mathbf{x}} $$
$$ = \frac{1}{4\pi\epsilon_0} \left( 2 \sigma \right) \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} d\phi \;\hat{\mathbf{x}} $$
$$ = \frac{1}{4\pi\epsilon_0} \left( 2 \sigma \right) \left( \pi \right) = \frac{\sigma}{2 \epsilon_0} $$
As you can see, because of the geometry of the infinite sheet, the dependence on the distance from the sheet fell out of the equation (with no approximations, for the most part). Doing the calculation from first principles, we have obtained an equation for the electric field via an infitie plate that one would normally find a textbook.
If your question asked for the actual reason (and not how we know it), this entire derivation is a consequence from Coulomb's law. To ask why Coulomb's law is as it is, is outside the scope of this answer (and physics?).
A: It will be much simple if you use Gauss' law to prove it with only a few lines than this complicated way of mathematical manipulation
Drawing n enclosed cylindrical Gaussian surface  with 2 end cap surfaces A arranged to pierce the infinite sheet of charges perpendicularly. Since the ekectric field lines are perpendicular to the sheet of charges, and the enclosed cylinder Gaussian surface is also perpendicular to the sheet of charge, the electric field lines must also perpendicular to the 2 cap end surface areas A, it means that the electric field vector E and the differential area vector of the differential area delta A are parallel pointing toward the same x direction
So tge dit product EdA can be expressed as ( Ei)(dAi) EdA i*i=EdA(1) = EdA
Now you just plug the result in Gauss' law 's equation for a charge in an  enclosed surface, and take the integral of it as follows:
Let Eo be the permitivity constant
Eo integral of EdA= EoE integral dA = Qenc
where Qenc is the charge on the sheet of charges enclosed by the piercing cylindrical Gaussian surface =aA where a is charge density and A is surface area
Since dA =A ----> the integral result is   EoEA= Qenc
Since there are 2 surface areas A, EoE (A+A) Qenc= aA ----> E = aA/2AEo
E = a/2Eo.
Since there is not any variable representing distance r in the equation for the ekectric field's magnitude, the magnitude of the electric field of the infinite sheet of charges is independent of the dustance between the sheet of charges and any point in the electric field , and  both a and Eo are constant , therefore E = constant at at all points in the electric field
A: Consider a negatively charged plate and an electron at a small distance from it.
The plate repels the charge. The electrons in the plate that are closest to the free electron push in perpendicular direction and also push the most because they are closer than any other electrons in the plate.
The other charges are at a greater distance and push less, and also mostly sideways. Due to symmetry, only the components perpendicular to the plate remain. For every charge on one side of the electron, there is another charge on the opposing side.
If you move the electron away from the plate, the amount of charges that push less sideways increases (more of the plate's charge is "under" the electron) by just the right amount to make up for the greater distance. 
