Two metal balls have charges of 7.1 * 10-6 coulombs and 6.9 * 10-6 coulombs. They are 5.7 * 10-1 meters apart. What is the force of interaction between the two balls? (k = 9.0 * 109 newton·meters2/coulomb2)

By the Coulomb's Law:

F = k * Q1 * Q2 / r²

k = 9.0 * 10^9

Q1 = 7.1 * 10 ^ (-6) C, Q2 = 6.9 * 10^ (-6) C, r = 5.7 * 10^ (-1)

F = 9.0 * 10^9 * 7.1 * 10^ (-6) * 6.9 * 10^ (-6) / (5.7 * 10^ (-1)) ² =

= 9.0 * 7.1 * 6.9 * 10^ (-12) / 32.49 * 10^ (-2) =

440.91 * 10^ (-10) / 32.49 =

= 13.57 * 10^ (-10) = 1.357 * 10^ (-9) N (the force of interaction between the two balls)

The answer is 1.4 newtons