Answer: 6.07 N
Explanation:
According to Coulomb's Law:
[tex]F_{E}= K\frac{q_{1}.q_{2}}{d^{2}}[/tex]
Where:
[tex]F_{E}[/tex] is the electrostatic force
[tex]K=8.99(10)^{9} Nm^{2}/C^{2}[/tex] is the Coulomb's constant
[tex]q_{1}=2.4(10)^{-8} C[/tex] and [tex]q_{2}=1.8(10)^{-6} C[/tex] are the electric charges
[tex]d=0.008 m[/tex] is the separation distance between the charges
Solving:
[tex]F_{E}= 8.99(10)^{9} Nm^{2}/C^{2}\frac{(2.4(10)^{-8} C)(1.8(10)^{-6} C)}{(0.008 m)^{2}}[/tex]
[tex]F_{E}=6.07 N[/tex]
