Answer:
q = 2.65 10⁻⁶ C
Explanation:
For this exercise we use Coulomb's law
F =[tex]k \frac{q_1q_2}{r^2}[/tex]
In this case they indicate that the load is of equal magnitude
q₁ = q₂ = q
the force is attractive because the signs of the charges are opposite
F = [tex]k \ \frac{q^2}{r^2}[/tex]
q = [tex]\sqrt{\frac{F \ r^2}{k} }[/tex]
we calculate
q = [tex]\sqrt{\frac{0.7 \ 0.3^2 }{9 \ 10^9} }[/tex]
q = [tex]\sqrt{7 \ 10^{-12} }[/tex]Ra 7 10-12
q = 2.65 10⁻⁶ C
