Answer: 181.23 C
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}=4.5 \mu C=4.5(10)^{-6} C[/tex] is the charge of object 1
[tex]q_{2}=2.8 \mu C=2.8(10)^{-6} C[/tex] is the charge of object 2
[tex]d=2.5 cm \frac{1 m}{100 cm}=0.025 m[/tex] is the separation distance between the charges
[tex]F_{E}=8.99(10)^{9} Nm^{2}/C^{2}\frac{(4.5(10)^{-6} C)(2.8(10)^{-6} C)}{(0.025 m)^{2}}[/tex]
[tex]F_{E}=181.23 C[/tex]
