Two 4.0kg masses are 1.0m apart on a frictionless table. each has 1.0μc of charge. part a what is the magnitude of the electric force on one of the masses?
what is the initial acceleration of each mass if they are released and allowed to move?

Electrostatic force = (Constant) x (charge-1) x (charge-2) / (distance²)

For all-SI units, the constant is almost exactly 9 x 10⁹ meter/farad.

So the force (on each mass) is

                   (9 x 10⁹) x (1.0 x 10⁻⁶) x (1.0 x 10⁻⁶) / (1.0)²  = 

                                               (9 x 1 x 1) x 10⁻³  /  (1)  = 

                                                       9 x 10⁻³  newtons .

To find the acceleration, use Newton's second law of motion: 

                                                                  Force  =  (mass) x (acceleration)

Divide each side by (mass):           Acceleration  =  (force) / (mass)

                                                       Acceleration  =  (9 x10⁻³) N / (4.0kg)

                                                       Acceleration  =    0.00225 meter/second²

If the charges have opposite signs, then the acceleration of each mass is
0.00225 m/s² in the direction toward the other one when you letum go. 
The force on each one, and therefore the acceleration, grows constantly,
as the distance between them shrinks.

If both charges have the same sign, then the acceleration of each mass is
0.00225 m/s² in the direction away from the other one when you letum go. 
The force on each one, and therefore the acceleration, shrinks constantly,
as the distance between them grows.

Did you ever figue this