## $\nabla \cdot \mathbf {E} ={\frac {\rho }{\varepsilon _{0}}}$

Integral form: $\Phi_E = \frac{Q}{\varepsilon_0} = \unicode{x222F}\, \mathbf{E} \cdot \mathrm{d}\mathbf{A}$: The total of the electric flux out of a closed surface is equal to the charge enclosed divided by the electric constant.

Differential form: $\nabla \cdot \mathbf {E} ={\frac {\rho }{\varepsilon _{0}}}$: The divergence of the electric field is equal to the charge density divided by the electric constant.

$\varepsilon _{0}$ = 8.854187817 x 10^{-12} F ˙ m^{-1}
a.k.a electric constant or vacuum permittivity, permittivity of free space. It relates electric flux in volt-meters to charge.