To find the charge density of an ideal electric dipole centred at the origin, I can evaluate the divergence of $\vec{E}(\vec{x})$ which equates to evaluating the laplacian of the potential.
Working backwards, $$-4\pi\nabla\delta^3(\vec{x})=\nabla\left[\nabla\cdot\left(\nabla\frac{1}{x}\right)\right]=\nabla \left[\nabla\cdot\left(-\frac{\vec{x}}{x^3}\right)\right]=\nabla \times \left[\nabla\times\left(\frac{\vec{x}}{x^3}\right)\right]+\nabla^2\left(-\frac{\vec{x}}{x^3}\right)=\nabla^2\left(-\frac{\vec{x}}{x^3}\right).$$
Including the dipole moment $\vec{p}_{0}$, necessary constants and using product rule, $$\nabla^2\left(\frac{\vec{p}_{0}\cdot\vec{x}}{4\pi\epsilon_0 x^3}\right)=-\frac{-\vec{p}_{0}\cdot\nabla\delta^3(\vec{x})}{\epsilon_0}.$$
From above, it is clear that the charge density of the ideal electric dipole is $-\vec{p}_{0}\cdot\nabla\delta^3(\vec{x})$.
However, the electric field of the same dipole is given as $$\vec{E}(\vec{x})=\frac{1}{4\pi\epsilon_0}\left[\frac{3(\vec{p}_{0}\cdot\hat{x})\hat{x}-\vec{p}_{0}}{x^3}\right]-\frac{1}{3\epsilon_0}\vec{p}_{0} \delta^3(\vec{x}) .$$
Since the first part of the above equation corresponds directly to the potential, it is expected that the charge density obtained from that part is $-\vec{p}_{0}\cdot\nabla\delta^3(\vec{x})$. The problem now is that I still have to take into account the Dirac term which when combined with the charge density obtained from the first term, will not give the same charge density as obtained from the potential.
I can argue that away from the origin, the Dirac term is zero and there is no contribution from it towards the charge density. However, Maxwell's equations should be valid for all points in space. As such, I should consider the origin as well but this leads back to the problem posed.
How do I resolve this conundrum?
