Why the coulomb's law is valid only for point and static charges?
Is there is any definite reason?
[Physics] Coulomb’s Law- Why the Coulomb’s law is valid only for point and static charges
coulombs-lawelectrostaticsreference framesspecial-relativity
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I believe the problem with your proof is that packing an area with spheres only takes up about 75% of a volume, and this problem doesn't disappear when you shrink the spheres down to infinitesimal sizes. No matter how small you shrink the spheres, those same packing properties should hold.
Your proof relies on saying that the integral over a large surface is equivalent to adding up tiny infinitesimal spherical surfaces. But packing with spheres won't give 100% efficiency. So the volume of the small spheres won't add up to the volume inside of the larger surface, and the two integrals $\int V d\tau$ won't be equivalent because they span different volumes.
There's the same kind of problem with the surface integral. If you try to add two shapes together with a flush side, the electric field will be equal while the normals will be in opposite directions, so the flush surface cancels out and you're left with a surface integral of just the boundary of the two shapes. With spherical packing, the surfaces will not be flush, and so the surface integral of the larger shape will be different than for the smaller shape.
The book proves you can make a dent in the sphere and the integral should remain the same. Therefore, you can draw any shape you want around a point charge and the integral should remain the same. This is equivalent to saying you can move a point charge anywhere you want inside an arbitrary surface. Now you can use superposition. If the law holds for any point charge at any position inside a surface, then break up your arbitrary charge distribution into infinitesimal point charges $dq$. The law holds for all $dq$, so it should hold for $\int_V dq d\tau$.
Proving a 'dent' in a sphere doesn't change the integral is equivalent to proving that for some arbitrary volume element, the integral is zero. This is because making a dent is equivalent to subtracting a small volume element. This is actually not too difficult. We can find $V$ just by inspection using $V=-\nabla E$. $$ V= \frac{q}{4\pi \epsilon_0 k}e^{\frac{-k}{\lambda}}$$ Convert the surface integral of the electric field into a volume integral with the divergence theorem. The field only depends on $r$ so using the spherical form of $\nabla$ we get $$ \nabla \cdot \vec{E }= \frac{1}{k^2}\frac{\partial}{\partial k} (k^2\vec{E})= \frac{1}{k^2}\frac{q}{4\pi \epsilon_0}\left( \frac{1}{\lambda} + (1 +\frac{k}{\lambda})\frac{-1}{\lambda}\right)e^{\frac{-k}{\lambda}}$$ $$ = \frac{q}{k^24\pi \epsilon_0} \frac{-k}{\lambda^2}e^{\frac{-k}{\lambda}}$$. Now simply rewrite the new Gauss's law. $$ \int_V \nabla \cdot \vec{E} d\tau + \frac{1}{\lambda^2}\int_V V d\tau= \frac{q}{4\pi \epsilon_0 \lambda^2}\int_V \left( \frac{-1}{k}e^{\frac{-k}{\lambda}} + \frac{1}{k}e^{\frac{-k}{\lambda}}\right)d\tau.$$
At $k=0$ the equation blows up on account of the point charge. For other finite $k$ values, this equation is just zero. This means that for any volume element that does not include the point charge, the two integrals add to zero. This means you can add or subtract any crazy surfaces you want from your original sphere as long as they don't include the origin.
Best Answer
It is really valid for distributions too, but you need to use the integral form.
To see why it is only "valid" for point charges, take a look at the equation (for a point particle, the one you have probably been looking at). It refers to the distance from the source, which is only defined for a point, not a distribution.
It is however not valid for moving charges. This is because the information about the position of the charge (the field caused by the charge) can only travel at the speed of light. See this link if you're interested in moving charges.