The Force on a Dipole p positioned between to point charges

Problem:  A dipole p is oriented along the x axis ((OverVector[p]) = - p (OverHat[x]);  p >0) between two point charges -q1 and + q2 (where q1>0 and q2 >0 are the magnitude of the charges:  |q1| and |q2|) each a distance s from the dipole as shown here:
						

Find the force on (OverVector[p])

(a) In an input cell write down an equation for the amplitude of the E field vs. x along the x axis (treat as a 1D problem) for points on the axis near x = 0.

(For the geometry shown, we agree that the field due to the two point charges along the x axis is given by:  (OverVector[E])[x] = E[x] (OverHat[x]); i.e., depends only on x and points in either the + or - x direction.  E[x] is the amplitude of the vector  (OverVector[E])[x].

(*  Input Cell -- enter your code here  *)
ClearAll["`*"]   (* houseleaning -- keep it *)

(b) What direction does the E field point at the position of the dipole?

(Enter answer in this text cell)

(c)  Given the geometry of this problem write down the Force on the dipole due to these point charges.

(Enter answer in this text cell)

(d) Use M to find an equation for F

(*  input cell  *)

(e) Comment on the direction of the force  (OverVector[F]) on (OverVector[p]).

(Enter answer in this text cell)

(f) Examine the slopes of EE[x] (by plotting EE[x] vs. x) near x = 0 For q1 > q2;  I take q1 = 2; and q2 = 1 for an example where q1 > q2.

Note that the Sign of the slope =  the Sign of (Subscript[Grad, x])(EE[x]  = Sign of (∂_x) EE[x]

(*  input cell; enter code here  *)

(Enter answer in this text cell)

(g) Deduce the direction of (OverVector[F]) on (OverVector[p]) from the slope of EE[x] at x = 0.

(Enter answer in this text cell)

(h)  Repeat  (f) and (g) for   q1 = 1; q2 = 2;   i.e., for q1< q2

(*  Enter Code Here  *)

(Enter answer in this text cell)

(i)  Using Manipulate, show the changes in slope (~Grad[EE[x] =  (∂_x)EE[x]) and resulting force on the dipole (~ - Grad[EE[x] = - (∂_x)EE[x]) due to various q1 and q2

I.e., Display {A Plot of EE[x], Resulting Force [should be a number]} while manipulating q1 and q2.

Clear[q1, q2]   (*  clean up  *)
 k=1; s =1;  (*  suggested values  *)

(*  A very simple two parameter Manipulate example is shown below  *)

Summary:  

The SIGN of the Slope of EE[x] at x = 0 determines direction of force.  (remember the - sign:  F = - Grad[p.E])

i.e., if q1 > q2, (∂_x) EE[x] is positive, F is in the -x direction;  If q1 < q2, (∂_x) EE[x]  is negative, F is in the +x direction.  Note the shape of the curve when for a given q1 you sweep q2 past q2 = q1; i.e., note the SLOPE of the curve at the position of the dipole (x = 0), as you do so.

The magnitude of the force ~ (1/(s^3))

Manipulate[Plot[q1 (x^2) - q2 (x^3), {x, 0,1}],{q1,1,4},{q2,0,10}]  (*  simple example  *)