Answer:
d) It will be cut to a fourth of the original force.
Explanation:
The magnitude of the electrostatic force between the charged objects is
[tex]F=k\frac{q_1 q_2}{r^2}[/tex]
where
k is the Coulomb's constant
q1 and q2 are the charges of the two objects
r is the separation between the two objects
In this problem, the initial distance is doubled, so
r' = 2r
Therefore, the new electrostatic force will be
[tex]F=k\frac{q_1 q_2}{(r')^2}=k\frac{q_1 q_2}{(2r)^2}=\frac{1}{4}(k\frac{q_1 q_2}{r^2})=\frac{1}{4}F[/tex]
So, the force will be cut to 1/4 of the original value.
