Answer:
if Q = 1c
[tex]F_{e} = 7.25 N[/tex]
Explanation:
We suposse that the earth has a charge equal to Q, then the moon would have a charge equal to 0.273 Q.
The electrical force between two objects charged is described by the Coulomb's law
[tex]F_{e} = \frac{kQ_{1}Q_{2} }{r^{2} }[/tex]
Where k is the Coulomb's constant [tex]k = 8.9874x10^{9} \frac{Nm^{2} }{c^{2} }[/tex]
r is the distance between the earth and the moon
Then the electrical force would be:
[tex]F_{e} = \frac{(8.9874x10^{9}Q(0.273Q) )}{338400000}[/tex]
[tex]F_{e} = 7.25 Q^{2} N[/tex]
For estimating this force we can suppose Q = 1 C and then
[tex]F_{e} = 7.25 N[/tex]
