Answer:
The objects were 1.8m apart.
Explanation:
We will start stating the Coulomb's Law. It says that:
[tex]F_e=\frac{Kq_1q_2}{r^{2}}[/tex]
Where F_e is the electric force between the objects, q_1 and q_2 are the magnitude of the charge of the objects, r is the distance between them and K is the Coulomb's constant ([tex]K=8.9*10^{9} \frac{Nm^{2} }{C^{2} }[/tex] in vacuum). Solving for the distance r we have:
[tex]r=\sqrt{\frac{Kq_1q_2}{F_e} }[/tex]
Plugging the given values into this equation, we obtain:
[tex]r=\sqrt{\frac{(8.9*10^{9}\frac{Nm^{2} }{C^{2} })(2.56*10^{-6}C)(3.34*10^{-7}C)}{2.26*10^{-3}N}}=1.8m[/tex]
In words, the two charged objects were 1.8m apart.
Answer:
1.8 m
Explanation:
Using coulomb's Law
F = kqq'/r²..................... Equation 1
Where F = Force of repulsion, q = First charge, q' = Second charge, r = distance between both charge k = proportionality constant.
make r the subject of the equation
r = √(kqq'/F).................. Equation 2
Given: F = 2.26×10⁻³ N, q = 2.56×10⁻⁶ C, q' = 3.34×10⁻⁷ C
Constant: k = 9×10⁹ Nm²/C²
Substitute into equation 2
r = √ [(2.56×10⁻⁶×3.34×10⁻⁷×9×10⁹ )/2.26×10⁻³]
r = √3.405
r = 1.8 m
r = 1.8 m
