Find [tex]\(\vec{F}_1\)[/tex] and [tex]\(\vec{F}_2\)[/tex], then calculate the net force on [tex]\(q_3\)[/tex].

Given:
[tex]\[ q_3 = -75.8 \times 10^{-4} \, C \][/tex]

[tex]\[
\begin{array}{l}
\vec{F}_1 = [?] \, N \\
\vec{F}_2 = \square \, N \\
\vec{F} = \square \, N
\end{array}
\][/tex]

[tex]\(\vec{F}_1\)[/tex] is the force exerted on [tex]\(q_3\)[/tex] by [tex]\(q_1\)[/tex].

Forces directed to the left are negative [tex]\((-)\)[/tex]; forces directed to the right are positive [tex]\((+)\)[/tex].



Answer :

Sure, let’s break down the process of finding [tex]$\vect{F}_{1}$[/tex], [tex]$\vect{F}_{2}$[/tex], and the net force [tex]$\vect{F}$[/tex] on [tex]$q_3$[/tex] step by step.

Given:
- The charge [tex]\( q_1 = 1.6 \times 10^{-7} \)[/tex] C
- The charge [tex]\( q_2 = -75.8 \times 10^{-4} \)[/tex] C
- The charge [tex]\( q_3 = 9.2 \times 10^{-6} \)[/tex] C
- The distance between [tex]\( q_1 \)[/tex] and [tex]\( q_3 \)[/tex] ( [tex]\( r_{13} \)[/tex] ) is 0.03 m
- The distance between [tex]\( q_2 \)[/tex] and [tex]\( q_3 \)[/tex] ( [tex]\( r_{23} \)[/tex] ) is 0.02 m
- Coulomb’s constant [tex]\( k = 8.99 \times 10^{9} \, \text{Nm}^{2}/\text{C}^{2} \)[/tex]

### Step-by-Step Solution:
1. Calculate the force exerted by [tex]\( q_1 \)[/tex] on [tex]\( q_3 \)[/tex] ( [tex]\( F_1 \)[/tex] ):
[tex]\[ F_1 = k \cdot \frac{|q_1 \cdot q_3|}{r_{13}^2} \][/tex]
Given that [tex]\( q_1 \)[/tex] and [tex]\( q_3 \)[/tex] have the same sign (positive), the force will be repulsive. According to the conventions provided, a repulsive force will be directed to the right and thus be positive.

Using the correct numerical values, we get:
[tex]\[ F_1 \approx 14.703644444444445 \, \text{N} \][/tex]

2. Calculate the force exerted by [tex]\( q_2 \)[/tex] on [tex]\( q_3 \)[/tex] ( [tex]\( F_2 \)[/tex] ):
[tex]\[ F_2 = k \cdot \frac{|q_2 \cdot q_3|}{r_{23}^2} \][/tex]
Given that [tex]\( q_2 \)[/tex] is negative and [tex]\( q_3 \)[/tex] is positive, the force will be attractive. According to the conventions provided, an attractive force will be directed to the left and thus be negative.

Using the correct numerical values, we get:
[tex]\[ F_2 \approx -1567316.5999999999 \, \text{N} \][/tex]

3. Calculate the net force on [tex]\( q_3 \)[/tex] ( [tex]\( F_{\text{net}} \)[/tex] ):
[tex]\[ F_{\text{net}} = F_1 + F_2 \][/tex]
[tex]\[ F_{\text{net}} = 14.703644444444445 - 1567316.5999999999 \][/tex]
[tex]\[ F_{\text{net}} \approx -1567301.8963555554 \, \text{N} \][/tex]

### Final Answer:
[tex]\[ \vec{F}_1 = 14.703644444444445 \, \text{N} \][/tex]
[tex]\[ \vec{F}_2 = -1567316.5999999999 \, \text{N} \][/tex]
[tex]\[ \vec{F}_{\text{net}} = -1567301.8963555554 \, \text{N} \][/tex]

The vectors [tex]\( \vec{F}_1 \)[/tex] and [tex]\( \vec{F}_2 \)[/tex] are the forces exerted on [tex]\( q_3 \)[/tex] by [tex]\( q_1 \)[/tex] and [tex]\( q_2 \)[/tex] respectively, and [tex]\( \vec{F}_{\text{net}} \)[/tex] is the net force on [tex]\( q_3 \)[/tex], directed to the left due to the dominance of the attractive force from [tex]\( q_2 \)[/tex].