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