Answer:
1. 579 x 10 ^-22N
Explanation:
F = kq1q2/r^2
= 9.0 x 10^9 x 5.67 x 10^-18 x 3.79 x 10^-18/ (3.5 x 10^-2)^2
= 1. 579 x 10 ^-22N
The attraction force between two charges is [tex]1.55 \times 10^{-22} \;\rm N[/tex].
Coulomb's law is used to find out the force of attraction between two charges placed at a certain distance. This law states that, when two point charges are placed at a distance, the force of attraction between them is proportional to the product of both charges and inversely proportional to the square of the distance between them.
[tex]F \propto q_1q_2[/tex] and [tex]F \propto \dfrac{1}{r^2}[/tex]
[tex]F =K \dfrac {q_1q_2}{r^2}[/tex]
Where K is the coulombs constant whose value is 8.988×109 N⋅m2⋅C−2.
Given that charges q1 and q2 are 5.67 x 10-18 C and 3.79 x 10 "C. The distance r is 3.5 x 10 m. The force of attraction is given below.
[tex]F = 8.98\times 10^9 \dfrac {5.67 \times 10^{-18}\times 3.79 \times 10^{-18}}{(3.5\times 10^{-2})^2}[/tex]
[tex]F = 1.55\times 10^{-22} \;\rm N[/tex]
Hence we can conclude that the attraction force between two charges is [tex]1.55 \times 10^{-22} \;\rm N[/tex].
To know more about the coulombs law, follow the link given below.
https://brainly.com/question/506926.
