Given:
• q1 = -4.60 x 10⁻⁶ C.
,
• q2 = +3.75 x 10⁻⁶ C.
,
• q3 = +8.30 x 10⁻⁵ C.
,
• d12 = 0.283 m
,
• d23 = 0.200 m
,
• d13 = 0.200
Let's find the magnitude of the net force on q3.
Let's first find the force acting on q1 and q3:
[tex]\begin{gathered} F_{13}=\frac{kq_1q_2}{(d_{13})^2} \\ \\ F_{13}=\frac{9\times10^9*4.60\operatorname{\times}10^{-6}*8.30\operatorname{\times}10^{-5}}{0.200^2} \\ \\ F_{13}=85.9\text{ N} \end{gathered}[/tex]
Also, for the force acting on q2 and q3, we have:
[tex]\begin{gathered} F_{23}=\frac{kq_2q_3}{(d_{23})^2} \\ \\ F_{23}=\frac{9\times10^9*3.75\operatorname{\times}10^{-6}*8.30\operatorname{\times}10^{-5}}{0.200^2} \\ \\ F_{23}=70.03\text{ N} \end{gathered}[/tex]
Therefore, the magnitude of the net force on q3 will be:
[tex]\begin{gathered} F_{net}=\sqrt{(F_{13})^2+(F_{23})^2} \\ \\ F_{net}=\sqrt{85.9^2+70.3^2} \\ \\ F_{net}=\sqrt{7379.6+4904.4} \\ \\ F_{net}=110.99\approx111\text{ N} \end{gathered}[/tex]
Therefore, the magnitude of the net force on q3 is 111 N.
• ANSWER:
111 N
