PX284 - O5 - electrostatics

\vec{\nabla} \times \vec{E} = 0

this implies that $\vec{E}$ is conservative $⟹ \vec{E} = - \vec{\nabla} ψ$ , the gradient of the electrostatic potential

\vec{\nabla} \cdot \vec{E} = \vec{\nabla} \cdot (- \vec{\nabla} ψ) = - \nabla^{2} ψ = \frac{ρ}{ε_{0}}

\nabla^{2} ψ = - \frac{ρ}{ε_{0}}

\nabla^{2} ψ = 0