Question

Two electrons with a charge of magnitude 1.6×10-19 C in an atom are separated by 1.5×10-10 m, the typical size of an atom. What is the electric force between them?

Answer 1

Answer:

$1.02\cdot 10^{-8} N$, repulsive

Explanation:

The magnitude of the electric force between two charged particles is given by Coulomb's law:

$F=k\frac{q_1 q_2}{r^2}$

where:

$k=8.99\cdot 10^9 Nm^{-2}C^{-2}$ is the Coulomb's constant

$q_1, q_2$ are the two charges of the two particles

r is the separation between the two charges

The force is:

- repulsive if the two charges have  same  sign

- Attractive if the two charges have opposite signs

In this problem, we have two electrons, so:

$q_1=q_2=1.6\cdot 10^{-19}C$ is the magnitude of the two electrons

$r=1.5\cdot 10^{-10} m$ is their separation

Substituting into the formula, we find the electric force between them:

$F=(8.99\cdot 10^9)\frac{(1.6\cdot 10^{-19})^2}{(1.5\cdot 10^{-10})^2}=1.02\cdot 10^{-8} N$

And the force is repulsive, since the two electrons have same sign charge.