Answer:
[tex]F_2=1.3\times 10^{-7}\ N[/tex]
Explanation:
Given that,
Initial force, [tex]F=5.2\times 10^{-7}\ N[/tex]
distance, d = 2 m
New distance, d' = 4 m
We need to find the new force. The electrostatic force between two charges is given by :
[tex]F=k\dfrac{q_1q_2}{d^2}[/tex]
So,
[tex]\dfrac{F_1}{F_2}=(\dfrac{d_2}{d_1})^2\\\\\dfrac{F_1}{F_2}=(\dfrac{4}{2})^2\\\\\dfrac{F_1}{F_2}=4\\\\F_2=\dfrac{F_1}{4}\\\\F_2=\dfrac{5.2\times 10^{-7}}{4}\\\\F_2=1.3\times 10^{-7}\ N[/tex]
So, the new force is equal to [tex]1.3\times 10^{-7}\ N[/tex].
