Protons in an atomic nucleus are typically 10−15 m apart. what is the electric force (in n) of repulsion between nuclear protons? (enter the magnitude.)

The electric force is given by: 
 F = [ k*(q1)*(q2) ] / d^2 
 F = Electric force 
 k = Coulomb's constant 
 q1 = Charge of one proton 
 q2 = Charge of second proton 
 d = Distance between centers of mass 
 Values: 
 F = unknown 
 k = 8.98E 9 N-m^2/C^2 
 q1 = 1.6E-19 
 q2 = 1.6E-19 
 d = 1.0E-15 m 
 Insert values into F = [ k*(q1)*(q2) ] / d^2 
 F = [ (8.98E 9 N-m^2/C^2) * (1.6E-19) * (1.6E-19) ] / (1.0E-15 m)^2 
 F = 229.888 N
 answer
 the electric force of repulsion between nuclear protons is 229.888 N

If an object is moving or in motion