Question: finding 2-D scalar potential of vector field with parameter

I'm having trouble figuring out how to find the 2-D scalar potential of a 3-D vector field projected onto a plane.  The vector field is computed on the plane, coordinates (X,Y), as a vector function with arguments (X,Y,u), with 'u' being a parameter.

After invoking the VectorCalculus package and specifying

    SetCoordinates(cartesian[X,Y]),

I used the field’s X- and Y-components Fx and Fy (these are much simplified examples):

    Fx := (X,Y,u) -> 2*(X-u)/((X-u)^2 + Y^2)^(3/2)
    Fy := (X,Y,u) ->     2*Y/((X-u)^2 + Y^2)^(3/2)

to define the projected 2-D vector field VF with the instruction

    VF := (X,Y,u) -> VectorField( 2, <Fx(X,Y,u), Fy(X,Y,u)> )

Maple 12 accepted this fine.  The scalar potential for this field is

    sp = 1 / sqrt((X-u)^2 + Y^2)

but when asked to produce this, Maple instead gave an error:

> spVF := ScalarPotential( VF(X,Y,u) );
Error, (in VF) invalid input: too many and/or wrong type of arguments passed
to VectorCalculus:-VectorField; first unused argument is Vector(2, {(1) = (2*X-2*u)/((X-u)^2+Y^2)^(3/2), (2) = 2*Y/((X-u)^2+Y^2)^(3/2)}, attributes = [coords = cartesian[X, Y]])

Would someone point me to what I’m doing wrong?  How is this sort of thing supposed to be done?

Thanks.

- Dushan Mitrovich