Respuesta :
Answer:
7.12 mm
Explanation:
From coulomb's law,
F = kqq'/r².................... Equation 1
Where F = force, k = proportionality constant, q and q' = The two point charges, r = distance between the two charges.
Make r the subject of the equation,
r = √(kqq'/F).......................... Equation 2
Given: q = q' = 75.0 nC = 75×10⁻⁹ C, F = 1.00 N
Constant: k = 9.0×10⁹ Nm²/C².
Substitute into equation 2
r = √[ (75×10⁻⁹ )²9.0×10⁹/1]
r = 75×10⁻⁹.√(9.0×10⁹)
r = (75×10⁻⁹)(9.49×10⁴)
r = 711.75×10⁻⁵
r = 7.12×10⁻³ m
r = 7.12 mm
Hence the distance between the point charge = 7.12 mm
The two point charges must be at a distance = 7.12 mm
The electrostatic force due to point charges is given by the Coulomb's law which is represented by t he equation (1)
[tex]F = k \times q_1\times q_2 /r^2 \\[/tex]........(1)
Where F = magnitude of electrostatic force in N.
k = Coulomb's Constant = 9 [tex]\times[/tex] 10^9 N⋅m^2/C^−2
r = Distance between the point charges in m.
[tex]q_1[/tex] and [tex]q_2[/tex] are the magnitude of point charges in C.
Given that
[tex]q_1[/tex] =[tex]q_2[/tex] = 75.0 nC = [tex]75\times10^{-9}\; C[/tex]
Electrostatic Force F = 1 N
Putting the values in equation (1) we get
1 = 9[tex]\times[/tex] 10^9
solving for r gives us
r = 0.0007115 m [tex]\approx[/tex] 7.12 mm
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https://brainly.com/question/1698562
