Answer:
9.0 units
Explanation:
The magnitude of the electrostatic force between two charged objects is given by Coulomb's law:
[tex]F=k\frac{q_1 q_2}{r^2}[/tex]
where:
[tex]k=9\cdot 10^9 Nm^{-2}C^{-2}[/tex] is the Coulomb's constant
[tex]q_1, q_2[/tex] are the two charges of the two objects
r is the separation between the two charges
The force is:
- Attractive if the charges have opposite signs
- Repulsive if the charges have same sign
In this probleem, the initial force between the two objects is
F = 18.0 units
Then, the charge of object 1 is halved, so the new charge is
[tex]q_1'=\frac{q_1}{2}[/tex]
Therefore the new force is
[tex]F'=\frac{kq_1' q_2}{r^2}=\frac{k(q_1/2)q_2}{r^2}=\frac{1}{2}(\frac{kq_1 q_2}{r^2})=\frac{F}{2}[/tex]
So, the force will also halve: therefore, the new force will be 9.0 units.
