Gauss's law in integral form states that the charge Q enclosed by a surface S is a constant (epsilon0) times the flux of the electric field E outward through S. (The orientation of the surface S is outward.) Find the charge enclosed by the cube with vertices (plusminus2, plusminus2, plusminus2) if the electric field is E(x, y, z) = 2xi + 3yj + zk. Q = epsilon 0 The easiest way to do this problem is to apply the divergence theorem. When you do, you will rediscover the differential form of Gauss's law: The charge density rho is given by rho = epsilon0 middot E.