Answer:
The electrostatic force between the two would increase to approximately [tex]2.4[/tex] times the initial value.
Explanation:
By Coulomb's Law, the magnitude of the electrostatic force between two point charges is:
[tex]\displaystyle F = \frac{k\, q_{1}\, q_{2}}{r^{2}}[/tex],
Where:
In this question, [tex]q_{1}[/tex] becomes [tex](5.4\, q_{1})[/tex], [tex]q_{2}[/tex] becomes [tex](4.6\, q_{2})[/tex], and [tex]r[/tex] becomes [tex](3.2\, r)[/tex]. Thus, the magnitude of the electrostatic force between the two charges would become:
[tex]\begin{aligned} & \frac{k\, (5.4\, q_{1})\, (4.6\, q_{2})}{(3.2\, r)^{2}} \\ =\; & \frac{(5.4)\, (4.6)}{(3.2)^{2}}\, \left(\frac{k\, q_{1}\, q_{2}}{r^{2}}\right) \\ \approx \; & 2.4\, \left(\frac{k\, q_{1}\, q_{2}}{r^{2}}\right)\end{aligned}[/tex].
In other words, the magnitude of the electrostatic force between the two charges would become approximately [tex]2.4[/tex] times the initial value.
