# Why do we need Gauss' laws for electricity and magnetism?

I don't agree that you get that you obtain the Gauss law using the method proposed. What you obtain instead is $$\frac{\partial\nabla\cdot\mathbf{B}}{\partial t} = 0,\\ \frac{1}{c^2}\frac{\partial\nabla\cdot\mathbf{E}}{\partial t} + \mu_0\nabla\cdot\mathbf{J}= \frac{1}{c^2}\frac{\partial\nabla\cdot\mathbf{E}}{\partial t} - \mu_0\frac{\partial\rho}{\partial t}=0.$$ These equations give you only the rate of change of $$\nabla\cdot\mathbf{B}$$ and $$\nabla\cdot\mathbf{E}$$, but not their value, which needs to be defined by time integration and gives you the answer up to a position-dependent constant (whose time derivative is zero). E.g., the Gauss law for the electricity is given now by $$\nabla\cdot\mathbf{E}(\mathbf{r},t) = \frac{1}{\epsilon_0}\rho(\mathbf{r},t) +C(\mathbf{r}).$$ So we do need an additional constraint to specify function $$C(\mathbf{r})$$, i.e. the Gauss law, which in these terms can be written as: $$C(\mathbf{r}) =0.$$