Question

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)

Answer 1

The answer is 1.4 newtons <span />

Answer 2

Answer : The force of interaction between them is 1.35 N

Explanation :

Charge 1 $q_1=7.1\times 10^{-6}\ C$

Charge 2,  $q_2=6.9\times 10^{-6}\ C$  

Distance between charges, $d=5.7\times 10^{-1}\ m$

The force of interaction between them is given by using Coulomb's law. It is written as :

$F=k\dfrac{q_1q_2}{d^2}$

k is electrostatic constant

$F=9\times 10^9\times \dfrac{7.1\times 10^{-6}\ C\times 6.9\times 10^{-6}\ C}{(5.7\times 10^{-1}\ m)^2}$

$F=1.35\ N$

Hence, this is the required solution.