Answer:
F = 3.58 N
Explanation:
To solve this problem we apply Coulomb's law:
Two point charges (q1, q2) separated by a distance (d) exert a mutual force (F) whose magnitude is determined by the following formula:
[tex]F = \frac{k*q_{1}*q_{2}}{d^{2}}[/tex] Formula (1)
[tex]K=8.99*10^{9} \frac{N*m^{2} }{C^{2} }[/tex] : Coulomb constant
q1, q2: charge in Coulombs (C)
d: distance between charges (m)
We apply formula 1 for the two situations presented in the problem:
1) [tex]20 = k * \frac{q*q}{d^{2}} = k * \frac{q^{2}}{d^{2}}[/tex] Equation (1)
2) [tex]F = k *\frac{2,58q*2,58q}{(6.1d)^{2}} = \frac{k*2.58^{2}*q^{2}}{6.1^{2}*d^{2}}[/tex] Equation (2)
Solving the equation for d^2
[tex]d^{2}=\frac{k*q^{2}}{20}[/tex] Equation (3)
We replace d^2 from equation (3) in equation (2)
[tex]F = \frac{k*2.58^2*q^2}{6.1^2*\frac{k*q^2}{20}}[/tex] (We eliminate k and q^2)
[tex]F = \frac{2.58^2*20}{6.1^2} = \frac{6.6564*20}{37.21} = 3.58 N[/tex]
