Answer: [tex]5.214(10)^{11} N[/tex]
Explanation:
According to Coulomb's Law:
"The electrostatic force [tex]F_{E}[/tex] between two point charges [tex]q_{1}[/tex] and [tex]q_{2}[/tex] is proportional to the product of the charges and inversely proportional to the square of the distance [tex]d[/tex] that separates them, and has the direction of the line that joins them".
Mathematically this law is written as:
[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}=5.8 C[/tex] and [tex]q_{2}=6.4 C[/tex] are the electric charges
[tex]d=80 cm \frac{1 m}{100 cm}=0.8 m[/tex] is the separation distance between the charges
Solving:
[tex]F_{E}=8.99(10)^{9} Nm^{2}/C^{2}\frac{(5.8 C)(6.4 C)}{(0.8 m)^{2}}[/tex]
[tex]F_{E}=5.214(10)^{11} N[/tex]
