Question

two charges attract each other with a force of 1.2 * 10⁻⁶N. Calculate the force that would act between the two charges when distance is doubled, halved, trebled.

Answer 1

1. When the distance is doubled: $3\cdot 10^{-7}N$

The electrostatic force between two charges is given by:

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

where

k is the Coulomb's constant

q1 and q2 are the two charges

r is the distance between the two charges

The initial force between the two charges is $F=1.2\cdot 10^{-6}N$. In this part of the problem, the distance between the two charges is doubled, so we can write

$r'=2r$

And substituting into the formula, we find the new force:

$F'=k\frac{q_1 q_2}{r'^2}=k\frac{q_1 q_2}{(2r)^2}=\frac{1}{4}k\frac{q_1 q_2}{r^2}=\frac{F}{4}$

So, the force is reduced to 1/4 of its original value. Therefore, it is

$F'=\frac{1.2\cdot 10^{-6} N}{4}=3\cdot 10^{-7}N$

2. When the distance is halved: $4.8\cdot 10^{-6}N$

The initial force between the two charges is $F=1.2\cdot 10^{-6}N$. In this part of the problem, the distance between the two charges is halved, so we can write

$r'=r/2$

And substituting into the formula, we find the new force:

$F'=k\frac{q_1 q_2}{r'^2}=k\frac{q_1 q_2}{(r/2)^2}=4k\frac{q_1 q_2}{r^2}=4F$

So, the force is quadrupled. Therefore, it is

$F'=4(1.2\cdot 10^{-6} N)=4.8\cdot 10^{-6}N$

3. When the distance is tripled: $1.33\cdot 10^{-7}N$

The initial force between the two charges is $F=1.2\cdot 10^{-6}N$. In this part of the problem, the distance between the two charges is tripled, so we can write

$r'=3r$

And substituting into the formula, we find the new force:

$F'=k\frac{q_1 q_2}{r'^2}=k\frac{q_1 q_2}{(3r)^2}=\frac{1}{9}k\frac{q_1 q_2}{r^2}=\frac{F}{9}$

So, the force is reduced to 1/9 of its original value. Therefore, it is

$F'=\frac{1.2\cdot 10^{-6} N}{9}=1.33\cdot 10^{-7}N$