1. When the distance is doubled: ![3\cdot 10^{-7}N](https://tex.z-dn.net/?f=3%5Ccdot%2010%5E%7B-7%7DN)
The electrostatic force between two charges is given by:
![F=k\frac{q_1 q_2}{r^2}](https://tex.z-dn.net/?f=F%3Dk%5Cfrac%7Bq_1%20q_2%7D%7Br%5E2%7D)
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
. In this part of the problem, the distance between the two charges is doubled, so we can write
![r'=2r](https://tex.z-dn.net/?f=r%27%3D2r)
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}](https://tex.z-dn.net/?f=F%27%3Dk%5Cfrac%7Bq_1%20q_2%7D%7Br%27%5E2%7D%3Dk%5Cfrac%7Bq_1%20q_2%7D%7B%282r%29%5E2%7D%3D%5Cfrac%7B1%7D%7B4%7Dk%5Cfrac%7Bq_1%20q_2%7D%7Br%5E2%7D%3D%5Cfrac%7BF%7D%7B4%7D)
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](https://tex.z-dn.net/?f=F%27%3D%5Cfrac%7B1.2%5Ccdot%2010%5E%7B-6%7D%20N%7D%7B4%7D%3D3%5Ccdot%2010%5E%7B-7%7DN)
2. When the distance is halved: ![4.8\cdot 10^{-6}N](https://tex.z-dn.net/?f=4.8%5Ccdot%2010%5E%7B-6%7DN)
The initial force between the two charges is
. In this part of the problem, the distance between the two charges is halved, so we can write
![r'=r/2](https://tex.z-dn.net/?f=r%27%3Dr%2F2)
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](https://tex.z-dn.net/?f=F%27%3Dk%5Cfrac%7Bq_1%20q_2%7D%7Br%27%5E2%7D%3Dk%5Cfrac%7Bq_1%20q_2%7D%7B%28r%2F2%29%5E2%7D%3D4k%5Cfrac%7Bq_1%20q_2%7D%7Br%5E2%7D%3D4F)
So, the force is quadrupled. Therefore, it is
![F'=4(1.2\cdot 10^{-6} N)=4.8\cdot 10^{-6}N](https://tex.z-dn.net/?f=F%27%3D4%281.2%5Ccdot%2010%5E%7B-6%7D%20N%29%3D4.8%5Ccdot%2010%5E%7B-6%7DN)
3. When the distance is tripled: ![1.33\cdot 10^{-7}N](https://tex.z-dn.net/?f=1.33%5Ccdot%2010%5E%7B-7%7DN)
The initial force between the two charges is
. In this part of the problem, the distance between the two charges is tripled, so we can write
![r'=3r](https://tex.z-dn.net/?f=r%27%3D3r)
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}](https://tex.z-dn.net/?f=F%27%3Dk%5Cfrac%7Bq_1%20q_2%7D%7Br%27%5E2%7D%3Dk%5Cfrac%7Bq_1%20q_2%7D%7B%283r%29%5E2%7D%3D%5Cfrac%7B1%7D%7B9%7Dk%5Cfrac%7Bq_1%20q_2%7D%7Br%5E2%7D%3D%5Cfrac%7BF%7D%7B9%7D)
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](https://tex.z-dn.net/?f=F%27%3D%5Cfrac%7B1.2%5Ccdot%2010%5E%7B-6%7D%20N%7D%7B9%7D%3D1.33%5Ccdot%2010%5E%7B-7%7DN)