Answer:
216.0 units
Explanation:
The electrostatic force between two charged objects is given by:
[tex]F=k\frac{q_1 q_2}{r^2}[/tex]
where
k is the Coulomb's constant
q1 and q2 are the two charges
r is the distance between the charges
In this problem, the charge of object 2 is tripled, so
[tex]q_2 ' = 3 q_2[/tex]
Substituting into the original equation,
[tex]F'=k\frac{q_1 (3 q_2)}{r^2}=3 (k\frac{q_1 q_2}{r^2})=3 F[/tex]
So, the electrostatic force will be tripled.
Since the initial electrostatic force is
F = 72.0 units
the new force will be
[tex]F' = 3 F = 3(72.0 )=216.0[/tex]
