The initial force between the two charges is given by:
[tex]F=k \frac{q_1 q_2}{d^2}[/tex]
where k is the Coulomb's constant, q1 and q2 the two charges, d their separation. Let's analyze now the other situations:
1. F
In this case, q1 is halved, q2 is doubled, but the distance between the charges remains d.
So, we have:
[tex]q_1' = \frac{q_1}{2}\\q_2' = 2 q_2\\d' = d[/tex]
So, the new force is:
[tex]F'=k \frac{q_1' q_2'}{d'^2}= k \frac{(\frac{q_1}{2})(2q_2)}{d^2}=k \frac{q_1 q_2}{d^2}=F[/tex]
So the force has not changed.
2. F/4
In this case, q1 and q2 are unchanged. The distance between the charges is doubled to 2d.
So, we have:
[tex]q_1' = q_1\\q_2' = q_2\\d' = 2d[/tex]
So, the new force is:
[tex]F'=k \frac{q_1' q_2'}{d'^2}= k \frac{q_1 q_2)}{(2d)^2}=\frac{1}{4} k \frac{q_1 q_2}{d^2}=\frac{F}{4}[/tex]
So the force has decreased by a factor 4.
3. 6F
In this case, q1 is doubled and q2 is tripled. The distance between the charges remains d.
So, we have:
[tex]q_1' = 2 q_1\\q_2' = 3 q_2\\d' = d[/tex]
So, the new force is:
[tex]F'=k \frac{q_1' q_2'}{d'^2}= k \frac{(2 q_1)(3 q_2)}{d^2}=6 k \frac{q_1 q_2}{d^2}=6F[/tex]
So the force has increased by a factor 6.
Answer
F) q1 is halved, q2 is doubled, but the distance between the charges remains d.
F/4) q1 and q2 are unchanged. The distance between the charges is doubled to 2d.
6F) q1 is doubled and q2 is tripled. The distance between the charges remains d.
