Question

If a proton at a particular distance from a charged particle is repelled with a given force, by how much will the force decrease when the proton is 3 times farther away from the particle?

Answer 1

The force between the proton and a charged particle can be written as follows:

$$F=\dfrac{kq_1q_2}{r^2}.$$

If the proton is 3 times farther away from the particle the new force can be written as follows:

$$F_{new}=\dfrac{kq_1q_2}{(3r)^2}=\dfrac{1}{9}\dfrac{kq_1q_2}{r^2}=\dfrac{1}{9}F.$$

Therefore, the force decreases to $\dfrac{1}{9}$ its original value.