Question

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.

Short answer

Answer:

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

Step-by-step solution

Significance of the neutral atom

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.

Determination of the ratio of the force on the atom by the point charge change

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 $\mathbf{1}\mathbf{/}\mathbf{273}$.