Given:
• q1 = +4.60 x 10⁻⁶ C
,
• q2 = +3.75 x 10⁻⁶ C
,
• q3 = -8.30 x 10⁻⁵ C.
,
• d12 = 0.350 m
,
• d23 = 0.155 m
Let's find the magnitude of the net force on q2.
• First find the force acting on charge 1 and 2:
![\begin{gathered} F_(12)=-(kq_1q_2)/((d_(12))^2) \\ \\ F_(12)=-\frac{9*10^9*4.60\operatorname{*}10^(-6)*3.75\operatorname{*}10^(-6)}{0.350^2} \\ \\ F_(12)=-1.27\text{ N} \end{gathered}]()
• Let's find the force acting on charge 2 with respect to change 3:
![\begin{gathered} F_(23)=(kq_1q_2)/((d_(23))^2) \\ \\ F_(23)=\frac{9*10^9*3.75\operatorname{*}10^(-6)*8.30\operatorname{*}10^(-5)}{0.155^2} \\ \\ F_(23)=116.60\text{ N} \end{gathered}]()
• Now, for the magnitude of the net force on q2, we have:
Therefore, the magnitude of the force on q2 is 116.6 N.
• ANSWER:
116.6 N