Answer:
Approximately [tex]14\; {\rm N}[/tex].
Explanation:
By Coulomb's Law, the magnitude of the electrostatic force between two charges is proportional to the product of the magnitudes of the two charges.
For example, consider charges of magnitude [tex]q_{1}[/tex] and [tex]q_{2}[/tex] that are apart from one another by a distance of [tex]r[/tex] in between. Let [tex]k[/tex] denote Coulomb's constant. By Coulomb's Law, the magnitude of electrostatic force between the two charges would be:
[tex]\displaystyle F = \frac{k\, q_{1}\, q_{2}}{r^{2}}[/tex].
In this question, the product of the magnitude of the two charges was originally [tex]3\; {\rm \mu C} \times 7\; {\rm \mu C} = 21\; {\rm (\mu C)^{2}}[/tex]. After [tex](-1\; {\rm \mu C})[/tex] is added to each charge, product of the magnitude of the two charges would become [tex](3 - 1)\; {\rm \mu C} \times (7 - 1)\; {\rm \mu C} = 12\; {\rm (\mu C)^{2}}[/tex].
Thus, the product of the magnitude of the two charges has been scaled to [tex]12\; {\rm \mu C}[/tex] from [tex]21\; {\rm \mu C}[/tex] . The magnitude of the electrostatic force between the two charges would be scaled from [tex]25\; {\rm N}[/tex] to [tex]25\; {\rm N} \times (12 / 21) \approx 14\; {\rm N}[/tex].
