Certainly! Let's break down the problem step-by-step as requested:
Given:
- Charge [tex]\( q_1 = q_2 = q_3 = 4.44 \times 10^{-6} \)[/tex] Coulombs.
- Coulomb's constant [tex]\( k = 8.99 \times 10^9 \)[/tex] N·m²/C².
- Distance between [tex]\( q_1 \)[/tex] and [tex]\( q_2 \)[/tex] [tex]\( r_{12} = 1 \)[/tex] meter.
- Distance between [tex]\( q_1 \)[/tex] and [tex]\( q_3 \)[/tex] [tex]\( r_{13} = 1 \)[/tex] meter.
We need to find:
1. The force [tex]\( \overrightarrow{F_2} \)[/tex] exerted on [tex]\( q_1 \)[/tex] by [tex]\( q_2 \)[/tex].
2. The force [tex]\( \overrightarrow{F_3} \)[/tex] exerted on [tex]\( q_1 \)[/tex] by [tex]\( q_3 \)[/tex].
3. The net force on [tex]\( q_1 \)[/tex], considering the direction of the forces.
### Step-by-Step Solution:
1. Calculate the force [tex]\( \overrightarrow{F_2} \)[/tex] exerted on [tex]\( q_1 \)[/tex] by [tex]\( q_2 \)[/tex]:
Using Coulomb's Law:
[tex]\[
F = k \frac{q_1 q_2}{r_{12}^2}
\][/tex]
Substituting the given values:
[tex]\[
F_2 = 8.99 \times 10^9 \times \frac{4.44 \times 10^{-6} \times 4.44 \times 10^{-6}}{1^2}
\][/tex]
[tex]\[
F_2 = 8.99 \times 10^9 \times 19.7136 \times 10^{-12}
\][/tex]
[tex]\[
F_2 = 0.177225264 \text{ N}
\][/tex]
2. Calculate the force [tex]\( \overrightarrow{F_3} \)[/tex] exerted on [tex]\( q_1 \)[/tex] by [tex]\( q_3 \)[/tex]:
Similarly using Coulomb's Law:
[tex]\[
F_3 = k \frac{q_1 q_3}{r_{13}^2}
\][/tex]
Substituting the given values:
[tex]\[
F_3 = 8.99 \times 10^9 \times \frac{4.44 \times 10^{-6} \times 4.44 \times 10^{-6}}{1^2}
\][/tex]
[tex]\[
F_3 = 8.99 \times 10^9 \times 19.7136 \times 10^{-12}
\][/tex]
[tex]\[
F_3 = 0.177225264 \text{ N}
\][/tex]
3. Determine the net force on [tex]\( q_1 \)[/tex]:
Since the forces [tex]\( \overrightarrow{F_2} \)[/tex] and [tex]\( \overrightarrow{F_3} \)[/tex] are equal in magnitude but are acting in opposite directions, we need to consider their directions to find the net force.
Given that forces directed left are negative and forces directed right are positive, we assume:
- [tex]\( \overrightarrow{F_2} \)[/tex] is directed to the left (negative direction).
- [tex]\( \overrightarrow{F_3} \)[/tex] is directed to the right (positive direction).
Net force [tex]\( \overrightarrow{F_{\text{net}}} \)[/tex] on [tex]\( q_1 \)[/tex]:
[tex]\[
F_{\text{net}} = F_3 - F_2
\][/tex]
Substituting the values:
[tex]\[
F_{\text{net}} = 0.177225264 - 0.177225264
\][/tex]
[tex]\[
F_{\text{net}} = 0 \text{ N}
\][/tex]
Hence, the forces are equal and opposite, resulting in a net force of zero newtons on [tex]\( q_1 \)[/tex].
### Final Answers:
- Force [tex]\( \overrightarrow{F_2} \)[/tex]: [tex]\( 0.177225264 \)[/tex] N.
- Force [tex]\( \overrightarrow{F_3} \)[/tex]: [tex]\( 0.177225264 \)[/tex] N.
- Net force on [tex]\( q_1 \)[/tex]: [tex]\( 0 \)[/tex] N.
