Answer:
New force, [tex]F'=\dfrac{F}{12}[/tex]
Explanation:
Given that, two point charges attract each other with an electric force of magnitude F. It is given by :
[tex]F=k\dfrac{q_1q_2}{r^2}[/tex]
If one charge is reduced to one-third its original value and the distance between the charges is doubled such that,
[tex]q_1'=\dfrac{q_1}{3}[/tex], [tex]r'=2r[/tex]
[tex]F'=k\dfrac{q_1'q_2'}{r'^2}[/tex]
[tex]F'=k\dfrac{(q_1/3)q_2}{(2r)^2}[/tex]
[tex]F'=\dfrac{F}{12}[/tex]
So, the electric force between them is reduced to (1/12). Hence, the correct option is (d).
