Two charged dust particles exert a force of 7.9 10-2 N on each other. What will be the force if they are moved so they are only one sixth as far apart?

Question

Two charged dust particles exert a force of 7.9  10-2 N on each other. What will be the force if they are moved so they are only one sixth as far apart?

kainui · Answer

Since force is based on this formula:
$$F=k\frac{ q1 q2}{ r^2 }$$
We can see that the only thing changing is the distance, everything else is constant. Now we can divide this equation by itself to get

$$\frac{ F_1 }{ F_2 }=\frac{ r_2^2 }{ r_1^2 }$$

And we know that r1 is 1, r2 is 1/6 and F1 is the original force! Solvable!