To solve this problem, we need to find the forces [tex]\(\vec{F}_1\)[/tex] and [tex]\(\vec{F}_3\)[/tex] exerted on [tex]\(q_2\)[/tex] and then calculate the net force acting on [tex]\(q_2\)[/tex].
Given:
- Charge [tex]\(q_1\)[/tex] is [tex]\( -28.1 \times 10^{-6} \)[/tex] C.
- Force [tex]\(\vec{F}_1\)[/tex] exerted by [tex]\(q_1\)[/tex] on [tex]\(q_2\)[/tex] is [tex]\(-71.6\)[/tex] N.
- The distance between [tex]\(q_2\)[/tex] and [tex]\( q_3 \)[/tex] is [tex]\(0.3\)[/tex] meters.
- Charge [tex]\(q_3\)[/tex] is [tex]\( 28.1 \times 10^{-6} \)[/tex] C (identical in magnitude but opposite in sign to [tex]\(q_1\)[/tex]).
- Coulomb's constant, [tex]\(k\)[/tex], is [tex]\(8.99 \times 10^9\)[/tex] N m[tex]\(^2\)[/tex] C[tex]\(^{-2}\)[/tex].
### Step-by-Step Solution:
1. Given Force [tex]\(\vec{F}_1\)[/tex]:
- The force [tex]\(\vec{F}_1\)[/tex] exerted by [tex]\(q_1\)[/tex] on [tex]\(q_2\)[/tex] is provided as [tex]\(-71.6\)[/tex] N (directed to the left).
2. Calculate Force [tex]\(\vec{F}_3\)[/tex]:
- [tex]\(\vec{F}_3\)[/tex] is the force exerted by [tex]\(q_3\)[/tex] on [tex]\(q_2\)[/tex].
- Using Coulomb's law:
[tex]\[
\vec{F}_3 = k \frac{|q_3 \cdot q_1|}{d^2}
\][/tex]
- [tex]\(k = 8.99 \times 10^9\)[/tex] N m[tex]\(^2\)[/tex] C[tex]\(^{-2}\)[/tex]
- [tex]\(|q_3| = 28.1 \times 10^{-6}\)[/tex] C
- [tex]\(|q_1| = 28.1 \times 10^{-6}\)[/tex] C
- [tex]\(d = 0.3\)[/tex] meters
- Plugging in the given values, we get:
[tex]\[
\vec{F}_3 = 8.99 \times 10^9 \cdot \frac{(28.1 \times 10^{-6})^2}{0.3^2}
\][/tex]
- Evaluating the above expression provides [tex]\(\vec{F}_3 \approx -78.87\)[/tex] N (considering the direction of force, it is negative because it is directed to the left).
3. Calculate the Net Force on [tex]\(q_2\)[/tex]:
- The net force [tex]\(\vec{F}\)[/tex] acting on [tex]\(q_2\)[/tex] is the sum of [tex]\(\vec{F}_1\)[/tex] and [tex]\(\vec{F}_3\)[/tex]:
[tex]\[
\vec{F}_{net} = \vec{F}_1 + \vec{F}_3
\][/tex]
- Substituting the values:
[tex]\[
\vec{F}_{net} = -71.6 + (-78.87)
\][/tex]
[tex]\[
\vec{F}_{net} \approx -150.47 \, \text{N}
\][/tex]
### Final Answer
So the forces are:
- [tex]\(\vec{F}_1 = -71.6\)[/tex] N
- [tex]\(\vec{F}_3 \approx -78.87\)[/tex] N
The net force on [tex]\(q_2\)[/tex] is:
- [tex]\(\vec{F}_{net} \approx -150.47\)[/tex] N
Forces directed to the left are negative, hence both forces [tex]\(\vec{F}_1\)[/tex] and [tex]\(\vec{F}_3\)[/tex] are negative, and the net force is also negative, indicating it is directed to the left.
