Answer:
The new force is 16 times of the initial force.
Explanation:
The electric force between charges is given by :
[tex]F=\dfrac{kq_1q_2}{r^2}[/tex]
If the distance is halved, d' =d/2 and charges are doubles, [tex]q_1'=2q_1\ \text{and}\ q_2'=2q_2[/tex]
New force becomes,
[tex]F'=\dfrac{kq_1'q_2'}{r'^2}\\\\F'=\dfrac{k(2q_1)(2q_2)}{(d/2)^2}\\\\F'=16\times \dfrac{kq_1q_2}{r^2}\\\\F'=16F[/tex]
So, the new force is 16 times of the initial force.
If the distance is halved and the charges of both particles are doubled, the force is 16 times as great.
According to Coulomb's law, the electrostatic force between two charges is given by;
[tex]F=k\frac{q_1 q_2}{r^2}[/tex]
Where 'k' is the Coulomb's constant.
If the charges of both particles are doubled and the distance is halved, the new force will be;
[tex]F\,'=k\frac{2q_1 \times 2q_2}{(r/2)^2}=k \frac{4\times 4\times q_1 q_2}{r^2} =16\,F[/tex]
So, the new force will be 16 times greater than the old force.
Learn more about Coulomb's law here:
https://brainly.com/question/14049417
