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)=-(kq_1q_2)/((d_(12))^2) \\ \\ F_(12)=-(9*10^9*8.60*10^(-6)*5.10*10^(-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)=(kq_2q_3)/((d_(23))^2) \\ \\ F_(23)=(9*10^9*5.10*10^(-6)*(3.30*10^(-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)=√((-3.22)^2+6.305^2) \\ \\ F_(net)=√(10.38+39.75) \\ \\ F_(net)=√(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