Answer:
54.0 units
Explanation:
The electrostatic force between two charged particles is given by the equation:
[tex]F=k\frac{q_1 q_2}{r^2}[/tex]
where
k is the Coulomb constant
[tex]q_1, q_2[/tex] are the charges of the two particles
r is the separation between the particles
The direction of the force is:
- Repulsive if the two charges have same sign (++ or --)
- Attractive if the two charges have opposite sign (+-)
In this problem, the initial force is
F = 36.0 units
when we have charges [tex]q_1,q_2[/tex] and distance [tex]r[/tex].
Later:
- Charge of object 1 is halved, so
[tex]q_1'=\frac{q_1}{2}[/tex]
- Charge of object 2 is tripled, so
[tex]q_2'=3q_2[/tex]
So, the new electrostatic force will be:
[tex]F'=\frac{kq_1' q_2'}{r^2}=\frac{k(\frac{q_1}{2})(3q_2)}{r^2}=\frac{3}{2}(\frac{kq_1 q_2}{r^2})=\frac{3}{2}F[/tex]
And using F = 36.0, we find
[tex]F'=\frac{3}{2}(36.0)=54.0[/tex]
