Question

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

Answer 1

Answer:

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

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 ;

$F = \frac{Kq_1q_2}{r^2}$

where;

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

$F = \frac{Kq_1q_2}{r^2} \\\\F = \frac{(9\times 10^9)(0.06)(0.07)}{3^2} \\\\F = 4.2 \times 10^{6} \ N$

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