Question

find the net force on q₁. +4.44 μc +7.28 μc +2.25 μc q₁ q₂ q₃ ←0.100 m→ ←0.100 m→ \\(\vec{f}_2\\) = force exerted on q₁ by q₂ \\(\vec{f}_3\\) = force exerted on q₁ by q₃ \\(\vec{f}_2 = -29.1\\) n \\(\vec{f}_3 = ?\\) n \\(\sigma \vec{f} = \\) n remember: like charges repel; opposite charges attract.

Explanation:

Step1: Recall Coulomb's Law

$$F = k\frac{|q_1 q_2|}{r^2}$$
where $k=8.988\times10^9\ \text{N·m}^2/\text{C}^2$

Step2: Define values for $\vec{F_3}$

$q_1=4.44\times10^{-6}\ \text{C}$, $q_3=2.25\times10^{-6}\ \text{C}$, $r=0.200\ \text{m}$

Step3: Calculate magnitude of $\vec{F_3}$

$$|F_3| = 8.988\times10^9\times\frac{4.44\times10^{-6}\times2.25\times10^{-6}}{(0.200)^2} = 1.13\ \text{N}$$

Step4: Assign direction to $\vec{F_3}$

Like charges repel, so $\vec{F_3} = -1.13\ \text{N}$ (leftward, negative direction)

Step5: Calculate net force

$$\sum\vec{F} = \vec{F_2} + \vec{F_3} = -29.1 + (-1.13) = -30.23\ \text{N}$$

Answer:

$\vec{F_3} = -1.13\ \text{N}$, $\sum\vec{F} = -30.23\ \text{N}$