Answer :
Answer:
[tex]8 \times 10^{-9}\; {\rm N}[/tex].
Explanation:
By Coulomb's Law, the magnitude of electrostatic force between two point charges is:
[tex]\displaystyle F = \frac{k\, q_{1}\, q_{2}}{r^{2}}[/tex],
Where:
- [tex]k[/tex] is Coulomb's Constant,
- [tex]q_{1}[/tex] and [tex]q_{2}[/tex] are the magnitudes of the two charges, and
- [tex]r[/tex] is the distance between the two charges.
In this question, assume that the magnitude of the two point charges were originally [tex]q_{1}[/tex] and [tex]q_{2}[/tex] with a distance of [tex]r[/tex] in between.
Assume that [tex]q_{2}[/tex] becomes [tex](q_{2} / 2)[/tex] and [tex]r[/tex] becomes [tex](r / 2)[/tex]. By Coulomb's Law, the magnitude of the electrostatic force between the two new charges would become:
[tex]\begin{aligned}F &= \frac{k\, q_{1}\, (q_{2} / 2)}{(r / 2)^{2}} \\ &= \frac{k\, q_{1}\, q_{2} / 2}{r^{2} / 2^{2}} \\ &= \frac{2\, k\, q_{1}\, q_{2}}{r^{2}}\end{aligned}[/tex].
In other words, magnitude of the force between the two new charges would be twice that of the original value. The magnitude of the new force would be [tex]8 \times 10^{-9}\; {\rm N}[/tex].