Interaction energy between dipole and potential It is known that interaction energy = $-\vec{p}.\vec{E}$ where $\vec{p}$ is dipole moment and $\vec{E}$ is the electric field. 
I have to calculate the interaction energy of a system whose dipole moment and potential are the only available data. 
My assumption is since the negative gradient of potential is electric field, can we calculate the dipolar component of the potential as an alternative to electric field, i.e. 
$$\begin{align}
\phi_{dx}&=\phi_x(x-x_c) \\
\phi_{dy}&=\phi_y(y-y_c) \\
\phi_{dz}&=\phi_z(z-z_c)
\end{align}$$
where $x$, $y$, $z$ are the positions of atoms and $x_c$, $y_c$, $z_c$ is the geometrical centre, and then calculate the dot product with the dipole moment $(p_x,p_y,p_z)$.
Does it sounds meaningful?? Kindly provide some suggestions in solving my problem.
 A: What you're calling the "dipolar component of the potential" is not actually that object. For something to be a 'component' of the potential it also needs to be a scalar; in that sense, the sum of those three terms,
$$\Delta \phi=\sum_j\frac{\partial \phi}{\partial x_j}(x_j-x_{\text c,j})$$
could indeed be called the 'dipolar component', and it will work in any region where there are no charges other than test charges. You chould note in particular that this object is intimately related to the electric field, as
$$\Delta\phi=-\nabla \mathbf{E}\cdot(\mathbf{r}-\mathbf{r}_\text{c}).$$
Because of this, using the dipolar component will work as long as you evaluate it at the dipole separation.
On the other hand, if you take the vector 
$$\mathbf{\Delta\phi}=\sum_j\frac{\partial \phi}{\partial x_j}(x_j-x_{\text c,j})\mathbf{e}_j$$
and then dot that with the atoms' positions, then you will get an interaction energy
$$
U
=\sum_a\sum_j\frac{\partial \phi}{\partial x_j}(x_{a,j}-x_{\text c,j})p_j
=\frac2q\sum_j\frac{\partial \phi}{\partial x_j}p_j^2
$$
which is manifestly wrong.
So, bottom line: no.
A: The electric field is given by
$$
E=-\nabla V-\frac{\partial A}{\partial t},
$$
where $V$ is the electric potential and $A$ is the vector potential. So you can get the electric field from the potential. Or if you have measurements of a large enough set of points that are close together compared to the scales over which the field varies you might be able to calculate a good approximation to the field.
