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:
[tex]\begin{gathered} F_{12}=-\frac{kq_1q_2}{(d_{12})^2} \\ \\ F_{12}=-\frac{9\times10^9*4.60\operatorname{\times}10^{-6}*3.75\operatorname{\times}10^{-6}}{0.350^2} \\ \\ F_{12}=-1.27\text{ N} \end{gathered}[/tex]
• Let's find the force acting on charge 2 with respect to change 3:
[tex]\begin{gathered} F_{23}=\frac{kq_1q_2}{(d_{23})^2} \\ \\ F_{23}=\frac{9\times10^9*3.75\operatorname{\times}10^{-6}*8.30\operatorname{\times}10^{-5}}{0.155^2} \\ \\ F_{23}=116.60\text{ N} \end{gathered}[/tex]
• Now, for the magnitude of the net force on q2, we have:
[tex]\begin{gathered} F_{net}=\sqrt{(F_{12})^2+(F_{23})^2} \\ \\ F_{net}=\sqrt{(-1.27)^2+(116.60)^2} \\ \\ F_{net}=\sqrt{1.6129+13595.56} \\ \\ F_{net}=\sqrt{13597.1729} \\ \\ F_{nrt}=116.60\text{ N} \end{gathered}[/tex]
Therefore, the magnitude of the force on q2 is 116.6 N.
• ANSWER:
116.6 N
