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From Gauss's law we find that E in any point near an infinite charged metal plate is:

$$E = \frac{\sigma\epsilon_0}{2}$$

which indicates that $E$ is same for any distance, but it seems weird and untrue from the concept of Coulomb's law. If distance from the plate is decreased a charged particle should feel more force on it. Can you please explain?

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You can do the integral with Coulomb's law directly. More intuitively, you can consider that as you get closer, the different charge elements give a larger electric field, but they to be more parallel to the charged plane, and that component is cancelled out by the charge element on that opposite side.

The integral:

$$\vec E=\int \frac{k~dq}{r^2}\hat r=k\sigma \int\frac{dA}{r^2}\hat r $$

By symmetry, we can argue that the electric field will be perpendicular to the plane. I'll call that the $\hat z$ direction. Then we can find just the z component:

$$E_z=\vec E\cdot\hat r=k\sigma\int\frac{dA}{r^2}\hat r \cdot \hat z$$

This is best considered in cylindrical coordinates. If the point we're looking at is $d$ away from the plane, and $\rho$ is the distance from the point parallel to the plane, this comes out to:

$$ E_z=k\sigma\int\frac{dA}{r^2}\frac{d}{r}=k\sigma d\int_0^\infty\frac{2\pi\rho ~d\rho}{(\rho^2+d^2)^\left(\frac{3}{2}\right)}$$

Integration by u-substitution with $u=\rho^2+d^2$, $du=2\rho~d\rho$:

$$E_z=k\sigma\pi d\int_{d^2}^\infty \frac{du}{u^{\frac{3}{2}}}=k\sigma \pi d\left(\frac{2}{\sqrt{u}}\Big|_{d^2}^\infty\right)=2k\sigma\pi $$

Or in terms of $\epsilon_0$ instead of $k$: $$E_z=\frac{\sigma\epsilon_0}{2}$$

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It may sound weird, but if you think in the language of mathematics, you may get convinced. In this case, size of the plate (infinite) is much larger than any finite distance near the plate and that is why E is independent of any finite distance near the plate.
Chris has already nicely provided derivation in terms of Coulomb's law.

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