Answer:
E) d/sqrt2
Explanation:
The initial electric force between the two charge is given by:
[tex]F=k\frac{q_1 q_2}{d^2}[/tex]
where
k is the Coulomb's constant
q1, q2 are the two charges
d is the separation between the two charges
We can also rewrite it as
[tex]d=\sqrt{k\frac{q_1 q_2}{F}}[/tex]
So if we want to make the force F twice as strong,
F' = 2F
the new distance between the charges would be
[tex]d'=\sqrt{k\frac{q_1 q_2}{(2F)}}=\frac{1}{\sqrt{2}}\sqrt{k\frac{q_1 q_2}{(2F)}}=\frac{d}{\sqrt{2}}[/tex]
so the correct option is E.
