The force on Object A = 1116.07 N
Explanation:The separation between the two charges, d = 1.68 cm
d = 1.68/100 m
d = 0.0168 m
The charge on object A
[tex]\begin{gathered} q_A=5.0\mu C \\ q_A=5\times10^{-6}C \end{gathered}[/tex]The charge on object B
[tex]\begin{gathered} q_B=7.0\mu C \\ q_B=7\times10^{-6}C \end{gathered}[/tex]The electric constant
[tex]k=9\times10^9Nm^2C^{-2}[/tex]The force on on the charge A is calculated as shown below:
[tex]\begin{gathered} F_A=\frac{kq_Aq_B}{d^2} \\ F_A=\frac{9\times10^9\times5\times10^{-6}\times7\times10^{-6}}{0.0168^2} \\ F_A=\frac{0.315}{0.00028224} \\ F_A=1116.07N \end{gathered}[/tex]The force on Object A = 1116.07 N
