Explanation:
It is given that, an electron and a proton are separated by a distance of 1.0 m i.e d = 1 m . At this position, F is the force between them
[tex]F=k\dfrac{q_1q_2}{1^2}[/tex]...............(1)
We need to find effect on force between them if the electron moves 0.5 m away from the proton. Let the force is F'.
[tex]F'=k\dfrac{q_1q_2}{0.5^2}[/tex]...............(2)
On dividing equation (1) and (2) we get :
[tex]\dfrac{F}{F'}=\dfrac{k\dfrac{q_1q_2}{1^2}}{k\dfrac{q_1q_2}{0.5^2}}[/tex]
[tex]\dfrac{F}{F'}=0.25[/tex]
F' = 4 F
So, the distance between the electron and the proton is 0.5 m, the new force will be 4 times of previous force.