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

`1.02\cdot 10^(-8) N`, repulsive

Step-by-step explanation:

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

`F=k(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)((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.