1) A. 2
The electrostatic force between two 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 charges of the two objects
r is the separation between the two objects
In this problem, the charge of one object is doubled, so
[tex]q_1' = 2q_1[/tex]
Therefore the new force is
[tex]F'=k\frac{q_1' q_2}{r^2}=k\frac{(2q_1)q_2}{r^2}=2(\frac{kq_1 q_2}{r^2})=2F[/tex]
So, the force will double.
2) D. 1/4
Using the same formula, the electrostatic force between the two objects is:
[tex]F=k\frac{q_1 q_2}{r^2}[/tex]
In this problem, the distance between the two objects is doubled, so
[tex]r' = 2r[/tex]
Therefore the new force is
[tex]F'=k\frac{q_1 q_2}{(r')^2}=k\frac{q_1 q_2}{(2r)^2}=\frac{1}{4}(\frac{kq_1 q_2}{r^2})=\frac{F}{4}[/tex]
So, the force will decrease to 1/4 of its original value.
3) 4/9
Using the same formula, the electrostatic force between the two objects is:
[tex]F=k\frac{q_1 q_2}{r^2}[/tex]
In this problem, the amount of charge on both objects doubles, so
[tex]q_1' = 2q_1[/tex]
[tex]q_2' = 2q_2[/tex]
Also, the distance is tripled
[tex]r' = 3r[/tex]
Therefore the new force is
[tex]F'=k\frac{q_1' q_2'}{(r')^2}=k\frac{(2q_1) (2q_2)}{(3r)^2}=\frac{4}{9}(\frac{kq_1 q_2}{r^2})=\frac{4}{9}F[/tex]
So, the force will decrease to 4/9 of its original value.
