# Question #dbb08

Feb 16, 2016

$F = \frac{{Q}_{1} {Q}_{2}}{{r}^{2}}$

#### Explanation:

Coulomb had defined the electrostatic force by:

$F = \frac{{Q}_{1} {Q}_{2}}{{r}^{2}}$

where, ${Q}_{1} \mathmr{and} {Q}_{2}$ are the charges of the oppositely charged ions and $r$ is the distance between the two nuclei.

For example, when comparing the electrostatic forces in two different ionic compounds we consider the following:

$N a C l$ and $M g C {l}_{2}$:

We look at the difference in charges:

For $N a C l$, the product ${Q}_{1} {Q}_{2} = \left(+ 1\right) \times \left(- 1\right) = - 1$

For $M g C {l}_{2}$, the product ${Q}_{1} {Q}_{2} = \left(+ 2\right) \times \left(- 1\right) = - 2$

Which means that ${F}_{M g C {l}_{2}} > {F}_{N a C l}$.

$N a C l$ and $C s C l$:

We look at the difference of $r$.

Since $C {s}^{+}$ is greater in size than $N {a}^{+}$ therefore, ${r}_{C s C l} > {r}_{N a C l}$ and thus ${F}_{N a C l} > {F}_{C s C l}$.