$\text{From point form of Gauss’s Law} \\ ∇ . \bar{D}=ρ_v \\ ∇.(ε \bar{E})=ρ_v……\bar{D}= ε \bar{E} \\ \text{Reverse relation b/w ∇ and} \bar{E} is: \\ \bar{E}= -∇ v \\ \text{Taking divergence on both sides,} \\ ∇ . \bar{E}= -∇ (∇v) \\ \dfrac{ρ_v}{ε}= - ∇^2 v \\ ∇^2 v=\dfrac{-ρ_v}{ε} \ \ \ …