If the distance between a neutral atom and a point charge is tripled, by what factor does the force on the atom by the point charge change? Express your answer as a ratio: new force/original force.

When an atom has equal numbers of electrons and protons, the electric charge of that atom is zero, so it is said to be a neutral atom.

The equation of the original force between neutral atom by the point charge change is expressed as:

${\mathit{F}}_{\mathbf{1}}\mathbf{=}{\left(\frac{1}{4\pi {\epsilon }_0}\right)}^{\mathbf{2}}\left(\frac{2q^2\alpha }{r^5}\right)$

Here, $\mathbf{\left(}\frac{\mathbf{2}{\mathbf{q}}^{\mathbf{2}}\mathbf{\alpha }}{{\mathbf{r}}^{\mathbf{5}}}\mathbf{\right)}$ is the electric force constant, q is the charge of the point charge, $\mathit{\alpha }$ is the constant and r is the initial distance between the neutral atom and point charge.

The equation of the new force between neutral atom by the point charge change is expressed as:

${\mathit{F}}_{\mathbf{2}}\mathbf{=}{\mathbf{\left(}\frac{\mathbf{1}}{\mathbf{4}{\mathbf{\pi \epsilon }}_{\mathbf{0}}}\mathbf{\right)}}^{\mathbf{2}}\mathbf{\left(}\frac{\mathbf{2}{\mathbf{q}}^{\mathbf{2}}\mathbf{\alpha }}{{\mathbf{\left(}\mathbf{3}\mathbf{r}\mathbf{\right)}}^{\mathbf{5}}}\mathbf{\right)}$

Here, $\mathbf{\left(}\frac{\mathbf{1}}{\mathbf{4}{\mathbf{\pi \epsilon }}_{\mathbf{0}}}\mathbf{\right)}$ is the electric force constant, q is the charge of the point charge, $\mathit{\alpha }$ is the constant and r is the final distance between the neutral atom and point charge.

The ratio of the new force divided by the original force is calculated as:

$$\begin{gathered} \frac{{\mathbf{F}}_{\mathbf{2}}}{{\mathbf{F}}_{\mathbf{1}}}\mathbf{=}\frac{{\mathbf{\left(}\frac{\mathbf{1}}{\mathbf{4}{\mathbf{\pi \epsilon }}_{\mathbf{0}}}\mathbf{\right)}}^{\mathbf{2}}\mathbf{\left(}\frac{\mathbf{2}{\mathbf{q}}^{\mathbf{2}}\mathbf{\alpha }}{{\mathbf{\left(}\mathbf{3}\mathbf{r}\mathbf{\right)}}^{\mathbf{5}}}\mathbf{\right)}}{{\mathbf{\left(}\frac{\mathbf{1}}{\mathbf{4}{\mathbf{\pi \epsilon }}_{\mathbf{0}}}\mathbf{\right)}}^{\mathbf{2}}\mathbf{\left(}\frac{\mathbf{2}{\mathbf{q}}^{\mathbf{2}}\mathbf{\alpha }}{{\mathbf{r}}^{\mathbf{5}}}\mathbf{\right)}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{1}\mathbf{/}\mathbf{243}{\mathbf{r}}^{\mathbf{5}}}{\mathbf{1}\mathbf{/}{\mathbf{r}}^{\mathbf{5}}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{1}\mathbf{/}\mathbf{243} \end{gathered}$$

Thus, the force on the atom by the point charge increases by $\mathbf{1}\mathbf{/}\mathbf{273}$.