Question

If the distance between a neutral atom and a point charge is doubled, by what factor does the force on the atom by the point charge change?

Short answer

The factor by which force change is $\frac{1}{32}$.

Step-by-step solution

Significance of the force on the atom

The force on the atom is directly proportional to the square of the charge and the atomic polarizability and it is also inversely proportional to the five times of the distance.

Calculation of the force

The equation of the force on the atom for single point charge is expressed as:

$F_1={\left(\frac{1}{4\pi {\epsilon }_0}\right)}^2\frac{2q^2\alpha }{r^5}$

Here,$q$ is the charge of the proton, $\alpha$is the atomic polarizability,$\left(\frac{1}{4\pi {\epsilon }_0}\right)$is the force field constantand $r$is the distance between the neutral atom and the point charge.

The equation of the force on the atom for double point charge is expressed as:

$F_2={\left(\frac{1}{4\pi {\epsilon }_0}\right)}^2\frac{2q^2\alpha }{{\left(2r\right)}^5}$

Here,$q$ is the charge of the proton, $\alpha$is the atomic polarizability,$\left(\frac{1}{4\pi {\epsilon }_0}\right)$is the force field constantand $r$is the distance between the neutral atom and the point charge.

Ratio of single point charge force to doubled charge force is expressed as:

$$\begin{gathered} \frac{F_1}{F_2}=\frac{{\left({\displaystyle \frac{1}{4{\mathrm{\pi \epsilon }}_0}}\right)}^2{\displaystyle \frac{2q^2\alpha }{{\left(2r\right)}^5}}}{{\left({\displaystyle \frac{1}{4{\mathrm{\pi \epsilon }}_0}}\right)}^2{\displaystyle \frac{2q^2\alpha }{r^5}}} \\ =\frac{r^5}{32r^5} \\ =\frac{1}{32} \end{gathered}$$

Thus, the factor by which force change is $\frac{1}{32}$.