kale94
contestada

Two electrons are separated by a distance of 3.00 x 10^-6 meter. What are the magnitude and direction of the electrostatic forces each exerts on the other?

Respuesta :

skyluke89
The electrostatic force between two charges q1 and q2 is given by
[tex]F=k_e \frac{q_1 q_2}{r^2} [/tex]
where 
[tex]k_e = 8.99 \cdot 10^9 N m^2 C^{-2}[/tex] is the Coulomb's constant
[tex]r=3.00 \cdot 10^{-6} m[/tex] is the distance between the two charges.

In our problem, the two charges are two electrons, so their charges are equal and equal to 
[tex]q_1=q_2=q=-1.6 \cdot 10^{-19}C[/tex]

By substituting these values, we find the intensity of the force between the two electrons:
[tex]F=(8.99 \cdot 10^9 N m^2 C^{-2}) \frac{(-1.6 \cdot 10^{-19}C)(-1.6 \cdot 10^{-19}C)}{(3.00 \cdot 10^{-6} m)^2}=2.6 \cdot 10^{-17}N [/tex]

This is the magnitude of the force each electron exerts to the other one. The direction is given by the sign of the charges: since the two electrons have same charge, they repel each other, so the force exerted by electron 1 is toward electron 2 and viceversa.
onyebuchinnaji

The magnitude and direction of the electrostatic forces each exerts on the other are [tex]2.56 \times 10^{-17} \ N[/tex] in opposite direction.

The given parameters:

  • Distance between the two electrons, r = 3 x 10⁻⁶ m
  • Charge of each electron, q = 1.6 x 10⁻¹⁹ C.

The magnitude and direction of the electrostatic forces each exerts on the other is calculated as follows;

[tex]F = \frac{kq^2}{r^2} \\\\F = \frac{9\times 10^9 \times (1.6 \times 10^{-19})^2}{(3\times 10^{-6})^2} \\\\F = 2.56 \times 10^{-17} \ N[/tex]

Thus, the magnitude and direction of the electrostatic forces each exerts on the other are [tex]2.56 \times 10^{-17} \ N[/tex] in opposite direction.

Learn more about electrostatic force here: https://brainly.com/question/9774180

Otras preguntas