Question

Two identical ions are each missing two electrons, each ion has a charge of 2e. What is the magnitude of the force between the ions when their separation is 2.1×10−10m? k=8.99×109N⋅m2/C2, e=1.60×10−19C

Answer 1

Answer:

|F| = 2.09 × 10⁻⁸ assuming that the two ions are point charges.

Explanation:

What's the charge on each ion?

The symbol $e$ here stands for fundamental charge. Each electron carries a negative fundamental charge of -e. Each proton carry a positive fundamental charge of +e.

Molecules and atoms are neutral. They contain an equal number of electrons and protons. Remove one electron from a molecule or atom, and that particle will end up with more protons (which are positive) than electrons. That particle will carry a positive charge of +e become an ion (a cation to be precise.) Remove another electron and the ion will carry a charge of +2e.

For each ion $q = +2 \;e = 2\times 1.60\times 10^{-19}\;\text{C} = 3.2\times 10^{-19}\;\text{C}$.

What's the size of the electrostatic force between the two ions?

Consider Coulomb's Law for the electrostatic force $F$ between two point charges:

$\displaystyle F = -\frac{k\cdot q_1\cdot q_2}{r^{2}}$,

where

• $k$ is Coulomb's constant,
• $q_1$ and $q_2$ are the charge on the two point charges, and
• $r$ is the separation between the two charges.

Make sure that all values are in SI units. Assume that the two ions are small enough that they act like point charges:

$\displaystyle \begin{aligned}F &= -\frac{k\cdot q_1 \cdot q_2}{r^{2}}\\&=-\frac{8.99\times 10^{9}\cdot(3.2\times 10^{-19}) \cdot (3.2\times 10^{-19})}{(2.1\times 10^{-10})^{2}}\\ &= -2.09\times 10^{-8}\;\text{N}\end{aligned}$.

The value of $F$ is negative, meaning that the two charges will repel each other because they are both positive. The question is asking for the magnitude of this force. Thus drop the sign in front of $F$ to obtain $2.09\times 10^{-8}\;\text{N}$, which is the magnitude of $F$.