Given:
• Number of identical charges = 3
,
• Charge, q1 = q2 = q3 = 6.39 x 10⁻⁴ C.
,
• Length of each side of the triangle, d = 3.35 m
,
• θ = 60 degrees
Let's find the magnitude of the net force on q3.
For forces acting on q3 due to q1 and q2, apply the formula:
[tex]\begin{gathered} F=\frac{kq_1q_2}{d^2} \\ \\ F=\frac{9\times10^9*(6.39\times10^{-4})*(6.39\times10^{-4})}{3.35^2} \\ \\ F=\frac{3674.889}{11.2225} \\ \\ F=327.46\text{ N} \end{gathered}[/tex]
The magnitude of net force on all 3 charges will also be equal.
Now, to find the magnitude of the net force acting on q3, apply the formula:
[tex]\begin{gathered} F_{net}=\sqrt{F^2+F^2+2F^2cos\theta} \\ \\ F_{net}=\sqrt{2F^2+2F^2cos\theta} \end{gathered}[/tex]
Thus, we have:
[tex]\begin{gathered} F_{net}=\sqrt{2(327.46)^2+2(327.46)^2cos60} \\ \\ F_{net}=\sqrt{214460.1032+214460.1032(0.5)} \\ \\ F_{net}=\sqrt{214460.1032+107230.0516} \\ \\ F_{net}=\sqrt{321690.1548} \\ \\ F_{net}=567.18\text{ N} \end{gathered}[/tex]
Therefore, the magnitude of net force acting on charge 3 is 567.18 N.
• ANSWER:
567.18 N