Rediscutimos a validade da lei de Gauss no caso de distribuições discretas e contínuas de carga que se estendem por todo o espaço. Palavras-chave: eletromagnetismo clássico, eletrostática, lei de Gauss.We rediscuss the validity of Gauss's law in the case of homogenous discrete and continuous charge distributions fulfilling all space. Keywords: classical electromagnetism, electrostatics, Gauss's law.In physics, some questions that we can formulate though frequently not realizable from a practical point of view, must as a matter of principle be satisfactorily answered. Moreover, they can be pedagogical useful tools and help the teacher to introduce new topics or discuss particularly hard ones. The following interesting question belongs to this type of question and can be found in [1]: Given a uniform charge distribution ρ that fulfills all the space, what is the value of the electric field at an arbitrary point of the distribution? The answer given in [1] though correct is nevertheless intriguing. By symmetry the electric field must be zero everywhere which makes perfect sense, but this means that the electric flux through an arbitrary surface S enclosing a certain charge Q (S) is zero in conflict with the integral form of the Gauss's law that states that the flux must be equal to Q (S) / 0 . In order to keep the symmetrical solution for the field it is stated that Gauss's law is not valid in this situation because there are charges at the infinity. It is pedagogically worth to understand and rediscuss this problem and its answer and this is our aim in what follows.Let us begin by taking an alternative route to Gauss's law. Consider a point charge q placed at a distance d from a point C and a hypothetical Gaussian sphere of radius R centered at C, see Fig. 1. Consider also a point P on the surface of the Gaussian sphere. If the distance of the point charge to P is r, then the electric field at P is given by It is convenient to treat d, R, and r as vectors, then r = d + R and we can writêwhere θ plays the role of a polar angle. The electric flux through an infinitesimal area centered around P is given by d Φ = E·n da, where the outward normal vectorn at P can be set equal toê R = R/R. Therefore, the electric flux through the infinitesimal area da = R 2 sin θ dθ dϕ is