To determine the distance between a charged foam cup with a charge of [tex]\( 2.0 \times 10^{-6} \)[/tex] coulombs and a positively charged metal sphere with a charge of [tex]\( 2.5 \times 10^{-6} \)[/tex] coulombs, given that the electric force between them is 2.12 newtons, and using Coulomb's constant [tex]\( k = 9.0 \times 10^9 \)[/tex] newton-meter²/coulombs², we use Coulomb's law:
[tex]\[ F = k \frac{q_1 q_2}{r^2} \][/tex]
where:
- [tex]\( F \)[/tex] is the force between the charges,
- [tex]\( q_1 \)[/tex] and [tex]\( q_2 \)[/tex] are the magnitudes of the two charges,
- [tex]\( r \)[/tex] is the distance between the centers of the two charges.
Given:
- [tex]\( F = 2.12 \)[/tex] newtons,
- [tex]\( q_1 = 2.0 \times 10^{-6} \)[/tex] coulombs,
- [tex]\( q_2 = 2.5 \times 10^{-6} \)[/tex] coulombs,
- [tex]\( k = 9.0 \times 10^9 \)[/tex] newton-meter²/coulombs².
We need to find [tex]\( r \)[/tex], the distance between the charges.
Rearranging Coulomb's law to solve for [tex]\( r \)[/tex]:
[tex]\[ r^2 = k \frac{q_1 q_2}{F} \][/tex]
Substituting the given values:
[tex]\[ r^2 = 9.0 \times 10^9 \frac{(2.0 \times 10^{-6})(2.5 \times 10^{-6})}{2.12} \][/tex]
[tex]\[ r^2 = 9.0 \times 10^9 \frac{5.0 \times 10^{-12}}{2.12} \][/tex]
[tex]\[ r^2 = 9.0 \times 10^9 \times \frac{5.0 \times 10^{-12}}{2.12} \][/tex]
[tex]\[ r^2 = 9.0 \times 10^{-6} \times \frac{5.0}{2.12} \][/tex]
[tex]\[ r^2 = 9 \times 10^{3} \times 2.35849 \times 10^{-3} \][/tex]
[tex]\[ r^2 = 21.227 \times 10^{-6} \][/tex]
[tex]\[ r \approx \sqrt{0.022227} \][/tex]
[tex]\[ r \approx 0.1457 \][/tex]
The resulting distance is approximately 0.1457 meters, which corresponds most closely to option E: [tex]\( 1.5 \times 10^{-1} \)[/tex] meters.
Thus, the correct answer is:
E. [tex]\( 1.5 \times 10^{-1} \)[/tex] meters
