Question

A 06-C charge and a .07-C charge are apart at 3 m apart. What force attracts them?

Answer 1

the force of attraction between the two charges is 4.2 x 10⁹ N.

Step-by-step explanation:

Given;

the magnitude of first charge, q₁ = 0.06 C

the magnitude of the second charge, q₂ = 0.07 C

distance between the two charges, r = 3 m

The force of attraction between the two charges is calculated as ;

`F = (Kq_1q_2)/(r^2)`

where;

k is Coulomb's constant = 9 x 10⁹ Nm²/C²

`F = (Kq_1q_2)/(r^2) \\\\F = ((9* 10^9)(0.06)(0.07))/(3^2) \\\\F = 4.2 * 10^(6) \ N`

Therefore, the force of attraction between the two charges is 4.2 x 10⁹ N.