Given that the force is
[tex]F_1=\text{ 0.69 N}[/tex]The distance between the charged spheres is
[tex]r_1=\text{ 1.3 m}[/tex]We have to find the force when the distance is
[tex]\begin{gathered} r_2=\text{ 75 }cm\text{ } \\ =0.75\text{ m} \end{gathered}[/tex]The force is given by the formula
[tex]F=\frac{Kq1q2}{r^2}[/tex]Here K is Coulomb's constant.
q1 and q2 are the charges of the charged sphere.
Also,
[tex]\begin{gathered} kq1q2=Fr^2 \\ =F_1(r_1)^2 \\ =0.69\times(1.3)^2 \\ =1.1661Nm^2 \end{gathered}[/tex]The force when distance is r2 will be
[tex]\begin{gathered} F_2=\frac{kq1q2}{(r_2)^2} \\ =\text{ }\frac{1.1661}{(0.75)^2} \\ =2.1\text{ N} \end{gathered}[/tex]Thus, option d is correct.
