Answer:
[tex]F=-4496885.8N[/tex]
Explanation:
Use Coulomb's law, which states the electrical force with which two point resting charges are attracted or repelled is directly proportional to their product, inversely proportional to the square of the distance that separates them and acts in the direction of the line that joins them. In its scalar form, the law is:
[tex]F=k\frac{q_1*q_2}{r^2}[/tex]
Where:
[tex]k=Coulomb's\hspace{3}constant\approx9\times10^9\frac{N*m^2}{C}\\r=Distance\hspace{3}between\hspace{3}the\hspace{3}charges\\q_1\hspace{3}and\hspace{3}q_2= Signed\hspace{3}magnitudes\hspace{3}of \hspace{3}the \hspace{3}charges[/tex]
Express 1.70 km in meters:
[tex]1.70km*\frac{1000m}{1km} =1700m[/tex]
Finally, Replace the data provided by the problem in the previous equation:
[tex]F=(9\times10^9)\frac{(38)*(-38)}{(1700^2)} =-4496885.813\approx-4496885.8N[/tex]
