Answer:
D, it would decrease by a factor of 4
Explanation:
We know the force between two charged particles, also known as Coulombs law.
[tex]F = \frac{kq_{1}q_{2}}{r^2}[/tex]
We are only changing the distance, not the charges, so we can ignore both q and look at r, which is the distance between the two positively charged particles.
So let F be our initial condition:
[tex]F \alpha \frac{1}{r^2} \\F' \alpha \frac{1}{(2*r)^2} = \frac{1}{4*r^2}\\F' = \frac{1}{4} *F[/tex]
So the answer is D, it decreases by a factor of 4.
