Question

In the diagram, q₁ = +4.60 x 10-6 C,92 +3.75 x 10-6 C, and 93 = -8.30 x 10-5 C.Find the magnitude of the net force on 92.0.350 m9192↑0.155 m93(Make sure you know the direction of each force! Oppositesattract, similar repel.)

Answer 1

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:

$\begin{gathered} F_{12}=-\frac{kq_1q_2}{(d_{12})^2} \\ \\ F_{12}=-\frac{9\times10^9*4.60\operatorname{\times}10^{-6}*3.75\operatorname{\times}10^{-6}}{0.350^2} \\ \\ F_{12}=-1.27\text{ N} \end{gathered}$

• Let's find the force acting on charge 2 with respect to change 3:

$\begin{gathered} F_{23}=\frac{kq_1q_2}{(d_{23})^2} \\ \\ F_{23}=\frac{9\times10^9*3.75\operatorname{\times}10^{-6}*8.30\operatorname{\times}10^{-5}}{0.155^2} \\ \\ F_{23}=116.60\text{ N} \end{gathered}$

• Now, for the magnitude of the net force on q2, we have:

$\begin{gathered} F_{net}=\sqrt{(F_{12})^2+(F_{23})^2} \\ \\ F_{net}=\sqrt{(-1.27)^2+(116.60)^2} \\ \\ F_{net}=\sqrt{1.6129+13595.56} \\ \\ F_{net}=\sqrt{13597.1729} \\ \\ F_{nrt}=116.60\text{ N} \end{gathered}$

Therefore, the magnitude of the force on q2 is 116.6 N.

• ANSWER:

116.6 N