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

[tex]\[ -75.8 \times 10^{-6} C \][/tex]

[tex]\[
\begin{aligned}
\vec{F}_1 & = [?] \, \text{N} \\
\vec{F}_2 & = [?] \, \text{N} \\
\vec{F} & = [?] \, \text{N}
\end{aligned}
\][/tex]

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

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

Sagot :

GinnyAnswer
Sure, let's go through the details of the calculation step-by-step.

Given values:
- Charge \( q_1 = -75.8 \times 10^{-6} \) C
- Charge \( q_2 = 75.8 \times 10^{-6} \) C (same magnitude as \( q_1 \))
- Charge \( q_3 = 75.8 \times 10^{-6} \) C (same magnitude as \( q_1 \))

We also assume:
- The distance between \( q_1 \) and \( q_3 \) is \( 0.1 \) m
- The distance between \( q_2 \) and \( q_3 \) is \( 0.1 \) m

Coulomb's constant \( k = 8.99 \times 10^9 \, \text{N m}^2 / \text{C}^2 \).

### Step 1: Calculate Force \( F_1 \)
Force \( F_1 \) is the force exerted on \( q_3 \) by \( q_1 \).

Using Coulomb's law:
[tex]\[ F_1 = k \frac{|q_1 q_3|}{r^2} \][/tex]

Substitute the given values:
[tex]\[ F_1 = 8.99 \times 10^9 \, \frac{|(-75.8 \times 10^{-6}) \times (75.8 \times 10^{-6})|}{(0.1)^2} \][/tex]

Result:
[tex]\[ F_1 = 5165.33 \text{ N (rounding to 2 decimal places for simplicity)} \][/tex]

Since \( q_1 \) is negative and \( q_3 \) is positive, the force \( F_1 \) will be attractive, pulling \( q_3 \) to the left, hence:
[tex]\[ \vec{F}_1 = -5165.33 \text{ N} \][/tex]

### Step 2: Calculate Force \( F_2 \)
Force \( F_2 \) is the force exerted on \( q_3 \) by \( q_2 \).

Using Coulomb's law:
[tex]\[ F_2 = k \frac{|q_2 q_3|}{r^2} \][/tex]

Substitute the given values:
[tex]\[ F_2 = 8.99 \times 10^9 \, \frac{|(75.8 \times 10^{-6}) \times (75.8 \times 10^{-6})|}{(0.1)^2} \][/tex]

Result:
[tex]\[ F_2 = 5165.33 \text{ N (rounding to 2 decimal places for simplicity)} \][/tex]

Since both \( q_2 \) and \( q_3 \) are positive, the force \( F_2 \) will be repulsive, pushing \( q_3 \) to the right, hence:
[tex]\[ \vec{F}_2 = 5165.33 \text{ N} \][/tex]

### Step 3: Calculate Net Force \( F \) on \( q_3 \)

The net force \( \vec{F} \) is the sum of the forces \( \vec{F}_1 \) and \( \vec{F}_2 \):
[tex]\[ \vec{F} = \vec{F}_1 + \vec{F}_2 \][/tex]

Substitute the forces:
[tex]\[ \vec{F} = -5165.33 \text{ N} + 5165.33 \text{ N} \][/tex]

Result:
[tex]\[ \vec{F} = 10330.66 \text{ N} \][/tex]

### Summary
[tex]\[ \begin{aligned} \vec{F}_1 & = -5165.33 \text{ N (left)} \\ \vec{F}_2 & = 5165.33 \text{ N (right)} \\ \vec{F} & = 10330.66 \text{ N (net force to the right)} \\ \end{aligned} \][/tex]
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