Answer:
New force, [tex]F'=48\times 10^3\ N[/tex]
Explanation:
It is given that,
Force acting between two charged particles, [tex]F=7.5\times 10^2\ N[/tex]
We need to find the force if they are moved so they are only one-eighth as far apart.
The force between two charged particles separated at a distance of r is given by :
[tex]F=k\dfrac{q_1q_2}{r^2}[/tex]............(1)
If the charges are one-eighth as far apart then, r' =(1/8)r and new force is given by :
[tex]F'=k\dfrac{q_1q_2}{(\dfrac{r}{8})^2}[/tex]..........(2)
Dividing equation (1) and (2) :
[tex]\dfrac{F}{F'}=\dfrac{1}{64}[/tex]
[tex]F'=7.5\times 10^2\ N\times 64[/tex]
F' = 48000 N
or
[tex]F'=48\times 10^3\ N[/tex]
Hence, this is the required solution.
