Answer:
New force between them will become [tex]\frac{1}{36}[/tex] times
Explanation:
Let the charge on both the object are [tex]q_1\ and\ q_2[/tex] and distance between them is is given 1 cm
So r = 1 cm = 0.01 m
Electric force between them is given F
According to Coulomb's between two charges is given by
[tex]F=\frac{1}{4\pi \epsilon _0}\frac{q_1q_2}{r^2}=\frac{Kq_1q_2}{r^2}[/tex]
According to question [tex]F=\frac{Kq_1q_2}{0.01^2}=\frac{Kq_1q_2}{10^{-4}}[/tex]-----------eqn 1
Now distance is increased by 5 cm so new distance = 5+1 = 6 cm = 0.06 m
So new force [tex]F_{new}=\frac{Kq_1q_2}{0.06^2}=\frac{Kq_1q_2}{36\times 10^{-4}}[/tex]------------------eqn 2
Comparing eqn 1 and eqn 2
[tex]F_{new}=\frac{F}{36}[/tex]
