Question

1. what is the magnitude of electrostatic force between two electrons separated by a distance of 1.00×10⁻⁸ meters?

Explanation:

Step1: Identify the formula

Coulomb's law is $F = k\frac{q_1q_2}{r^2}$, where $k = 9.0\times10^{9}\ N\cdot m^{2}/C^{2}$, $q_1$ and $q_2$ are charges and $r$ is the distance between them. For two electrons, $q_1=q_2 = 1.6\times 10^{-19}\ C$.

Step2: Substitute the values

Substitute $k = 9.0\times10^{9}\ N\cdot m^{2}/C^{2}$, $q_1 = q_2=1.6\times 10^{-19}\ C$ and $r = 1.00\times 10^{-8}\ m$ into the formula: $F=9.0\times 10^{9}\times\frac{(1.6\times 10^{-19})\times(1.6\times 10^{-19})}{(1.00\times 10^{-8})^{2}}$.

Step3: Calculate the result

First, calculate the numerator: $(1.6\times 10^{-19})\times(1.6\times 10^{-19})=2.56\times 10^{-38}$. Then, $(1.00\times 10^{-8})^{2}=1.00\times 10^{-16}$. So $F = 9.0\times 10^{9}\times\frac{2.56\times 10^{-38}}{1.00\times 10^{-16}}$. $F=9.0\times2.56\times 10^{9 - 38+16}=23.04\times10^{-13}=2.304\times 10^{-12}\ N$.

Answer:

$2.304\times 10^{-12}\ N$