I find that we can test for the units of the Hartree potential by using a test molecule. I selected Li2 and applied an external electric field (Volts/Angstroms). First obtain the Hartree Potential cube file for a zero external potential
# electric field along the z direction at efieldmag eV/Angstrom or efieldmag V/Angstrom
&EXTERNAL_POTENTIAL
FUNCTION (A/B)*Z
VALUES [eV] ${efieldmag} [angstrom] 1.0
PARAMETERS A B
&END EXTERNAL_POTENTIAL
Cutoff = 400, RelCutoff=80
so we set efieldmag = 0.0
Now using a box/cell **without periodic boundary conditions** we can optimize the geometry ( an energy calculation should also work ) now we use cubecruncher to obtain the potential along the z axis. We expect the initial slope, which is equal to the initial electric field magnitude in the z-direction. Now molecules will have an electric field around them due to their charge distribution of nuclei and electrons so even when efieldmag=0.0 the electric field around the molecule will be nonzero and will diminish as you move away from the molecule. If the box is large enough the influence of the molecule on the initial electric field will nearly vanish far from the molecule. I used {Lx, Ly, Lz} of {12,12,15} , {12,12,20}, {12,12,25} and the initial slopes are -0.019047, -0.00027563, -0.0000032. The initial slope is obtained on the left in the limit as z --> 0. We see the influence of the molecule can be greatly reduced in a sufficiently large box and this is expected to be true for any molecule in a sufficiently large box/cell.
Now we set efieldmag=0.005 Volts/Angstrom, using a box with dimensions of {12,12,15}, {12,12,20}, {12,12, 25} and the initial slopes are -0.01932522, -0.00008413067, -0.0001806136
Next we set Cutoff=600, relCutoff=80 and box size {12,12, 25} and the initial slope is -0.0001811421
Next we set Cutoff=800, relCutoff=80 and box size {12,12, 25} and the initial slope is -0.0001815272
Next we set Cutoff=800, relCutoff=80 and box size {12,12, 35} and the initial slope is -0.0001837458
* assumption 1: Hartree potential has the units of hartree/e = 27.211386 Volts / a.u.
*assumption 2: Electric Field Magnitude (Volts/Angstroms) = 27.211386 * (Hartree Potential at a point)
Check: Given an electric field magnitude of 0.005 V/Ang we expect that the observed Hartree potential sufficiently far from the molecule will yield a slope along the z axis, Ez (electric field in the z direction) of the same magnitude. We expect that the result should approach (0.005 V/Ang)/27.211386 = 0.000183747 a.u. ( the expected slope of the Hartree potential due only to the external potential)
We see that the initial slope is equal to efieldmag/27.211386 (Volts/a.u.) which shows that assumption 1 holds.