Answer:The average separation between the charges is 1,951.73 m.
Explanation:
Let the charge on the near the top of the cloud is [tex]q_1[/tex]=43.2 C
Let the charge on the near the bottom of the cloud is [tex]q_2[/tex]=-38.7 C
Electric force between these two charges,F = [tex]3.95\times 10^{6} N[/tex]
Separation between the charges = r
[tex]F=k\times\frac{|q_1|\times |q_2|}{r^2}[/tex]
[tex]3.95\times 10^{6} N=9\times 10^9 N m^2/C^2\times \frac{43.2 C\times (38.7)}{r^2}[/tex]
[tex]r=1,951.73 m[/tex]
The average separation between the charges is 1,951.73 m.
This question involves the concept of Colomb's Law and electrostatic force.
The average separation between these charges is "1951.73 m".
According to Colomb's Law, every charge exerts an electrostatic force on the other charge, which is directly proportional to the product of the magnitudes of both the charges and inversely proportional to the square of the distance between them.
[tex]F=\frac{kq_1q_2}{r^2}[/tex]
where,
Therefore,
[tex]3.95\ x\ 10^6\ N=\frac{(9\ x\ 10^9\ N.m^2/C^2)(43.2\ C)(38.7\ C)}{r^2}\\\\r=\sqrt{\frac{(9\ x\ 10^9\ N.m^2/C^2)(43.2\ C)(38.7\ C)}{3.95\ x\ 10^6\ N}}[/tex]
r = 1951.73 m
Learn more about Colomb's Law here:
brainly.com/question/9774180
