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$\nabla \cdot \vec E(r) = \dfrac {1}{r^2} \dfrac {d(r^2 E)}{dr} \ne \dfrac{dE}{dr}$ in spherical coordinates.


The RHS side of Gauss's Law, that is the charge enclosed should remain the same is indeed true. The apparent confusion if any, should be in the LHS of the equation, the integral of the 'dot product' of field and area vectors. Consider the diagram, Now, when we take the dot product of the field vector with the area vector in the initial case, the field and ...


You are confusing chemistry and physics. In chemistry you may have learnt that when metals react, they give up electrons (the are oxidized). That is because there are usually a small number of electrons in the outermost orbit of the atom, and when these are released the atom is left with a very stable electron configuration. But when you are looking at a ...

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