Respuesta :
The force which acts on the third charge placed in between the two charges can be expressed as,
[tex]\begin{gathered} F=\frac{kq_1q_3}{r^2}+\frac{kq_{2_{}}_{}q_3}{r^2} \\ =\frac{kq_3}{r^2}(q_1+q_2) \end{gathered}[/tex]Here, F is the net force, k is the force constant, q1 is the first charge, q2 is the second charge, q3 is the second charge and r is the distance between first and third charge. Since the third charge is placed in midway of both the charges therefore,
[tex]\begin{gathered} r=\frac{1\text{ m}}{2} \\ =0.5\text{ m} \end{gathered}[/tex]Plug in the known values in the expression,
[tex]\begin{gathered} F=\frac{(9\times10^9Nm^2C^{-2})(20\text{ }\mu C)(\frac{10^{-6}\text{ C}}{1\text{ }\mu C})}{(0.5m)^2}(50\text{ }\mu C-75\text{ }\mu C) \\ =(720\times10^3\text{ N/C)(-}25\text{ }\mu C)(\frac{10^{-6}\text{ C}}{1\text{ }\mu C}) \\ =-18\text{ N} \end{gathered}[/tex]Therefore, the net force which acts on the third charge is -18 N in which negative sign indicates that the direction of force is towards the negative charge of the system.
