Question

Objects 1 and 2 attract each other with an electrostatic force of 18 units. If the charge of Object 2 is increased by a factor of 3 AND the distance separating Objects 1 and 2 is increased by a factor of 3, then the new electrostatic force will be _____ units.

Answer 1

Answer:

$F'=18/3=6\ units$

Explanation:

Electrostatic Force

We know particles 1 and 2 attract each other with an electrostatic force of 18 units. The force between two charged particles is given by Coulomb's formula

$\displaystyle F=\frac{k\ q_1\ q_2}{r^2}$

Where $q_1,\ q_2$ are the charges and r is the distance between them.

We are told the new charge 2 is

$q_2'=3q_2$

And the distance is three times the original

$d'=3d$

The new force is

$\displaystyle F'=\frac{k\ q_1\ (3q_2)}{(3r)^2}$

Operating

$\displaystyle F'=\frac{3k\ q_1\ q_2}{9r^2}$

$\displaystyle F'=\frac{k\ q_1\ q_2}{3r^2}$

This force is one third of the other one, so  

$\boxed{F'=18/3=6\ units}$