Answer:
2.30 × 10⁻⁸ N if the two electrons are in a vacuum.
Explanation:
The Coulomb's Law gives the size of the electrostatic force [tex]F[/tex] between two charged objects:
[tex]\displaystyle F = -\frac{k\cdot q_1 \cdot q_2}{r^{2}}[/tex],
where
- [tex]k[/tex] is coulomb's constant. [tex]k = 8.99\times 10^{8}\;\text{N}\cdot\text{m}^{2}\cdot\text{C}^{-2}[/tex] in vacuum.
- [tex]q_1[/tex] and [tex]q_2[/tex] are the signed charge of the objects.
- [tex]r[/tex] is the distance between the two objects.
For the two electrons:
- [tex]q_1 = q_2 = 1.60\times 10^{-19}\;\text{C}[/tex].
- [tex]r = 1\times 10^{-10}\;\text{m}[/tex].
- [tex]\displaystyle F = -\frac{k\cdot q_1 \cdot q_2}{r^{2}} = -\frac{8.99\times 10^{8}\times (1.60\times 10^{-19})^{2}}{(1\times 10^{-10})^{2}} = 2.30\times 10^{-8}\;\text{N}[/tex].
The sign of [tex]F[/tex] is negative. In other words, the two electrons repel each other since the signs of their charges are the same.
