Answer:
F'=(8/3)F
Explanation:
to find the change in the force you take into account that the electric force is given by:
[tex]F=k\frac{q_1q_2}{(18.0u)^2}=k\frac{q_1q_2}{324.0}u^2[/tex]
However, if q1'=1/3*q, q2'=2*q2 and the distance is halved, that is 18/=9.o unit:
[tex]F'=k\frac{q_1'q_2'}{(9u)^2}=k\frac{(1/3)q_1(2)q_2}{81.0u^2}=\frac{2}{3}k\frac{q_1q_2}{81.0u^2}[/tex]
if you multiply this result by 4 and divide by 4 you get:
[tex]F'=\frac{8}{3}k\frac{q_1q_2}{324.0u^2}=\frac{8}{3}F[/tex]
hence, the new force is 8/3 of the previous force F.