Question

Scientists experimenting with two charged objects of 2.56 x 10-6 C and 3.34 x 10-7 C displayed a repulsion of 2.26 x 10-3 N. How far apart were the two charged objects during the experiment to the nearest tenth of a meter?

Answer 1

Answer:

The objects were 1.8m apart.

Explanation:

We will start stating the Coulomb's Law. It says that:

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

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 ($K=8.9*10^{9} \frac{Nm^{2} }{C^{2} }$ in vacuum). Solving for the distance r we have:

$r=\sqrt{\frac{Kq_1q_2}{F_e} }$

Plugging the given values into this equation, we obtain:

$r=\sqrt{\frac{(8.9*10^{9}\frac{Nm^{2} }{C^{2} })(2.56*10^{-6}C)(3.34*10^{-7}C)}{2.26*10^{-3}N}}=1.8m$

In words, the two charged objects were 1.8m apart.

Answer 2

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