Question

In the diagram, q1 = +8.60 x 10-6 C,92 +5.10 x 10-6 C, and q3 = -3.30 x 10-6 C.Find the magnitude of the net force on 92.0.350 m9192magnitude (N)0.155 m93(Make sure you know the direction of each force!Opposites attract, similar repel.)

Answer 1

Given:

• q1 = +8.60 x 10⁻⁶ C

,

• q2 = +5.10 x 10⁻⁶ C.

,

• q3 = -3.30 x 10⁻⁶ C.

,

• d12 = 0.350 m

,

• d23 = 0.155m

Let's find the magnitude of the net force on q2.

First find the force on the charge 1 and 2:

$\begin{gathered} F_{12}=-\frac{kq_1q_2}{(d_{12})^2} \\ \\ F_{12}=-\frac{9\times10^9*8.60\times10^{-6}*5.10\times10^{-6}}{0.350^2} \\ \\ F_{12}=-3.22\text{ N} \end{gathered}$

Also for the force on charge 2 and 3, we have:

$\begin{gathered} F_{23}=\frac{kq_2q_3}{(d_{23})^2} \\ \\ F_{23}=\frac{9\times10^9*5.10\times10^{-6}*(3.30\times10^{-6})}{0.155^2} \\ \\ F_{23}=6.305\text{ N} \end{gathered}$

Now, the magnitude of net force on q2 will be:

$\begin{gathered} F_{net}=\sqrt{(F_{12})^2+(F_{23})^2} \\ \\ F_{net}=\sqrt{(-3.22)^2+6.305^2} \\ \\ F_{net}=\sqrt{10.38+39.75} \\ \\ F_{net}=\sqrt{50.13} \\ \\ F_{net}=7.08\text{ N} \end{gathered}$

Therefore, the magnitude of the net force on q2 is 7.08 N.

• ANSWER:

7.08 N