To solve this problem, we'll calculate the forces exerted on charge $q_3$ by both $q_1$ and $q_2$ using Coulomb's law, and then find the vector sum of these forces.

Step 1: Coulomb’s Law

Coulomb's law is given by: $F = k \frac{|q_1 q_2|}{r^2}$ where:

• $F$ is the magnitude of the force between the charges,
• $k = 8.99 \times 10^9 \, \text{N}\cdot\text{m}^2/\text{C}^2$ is Coulomb's constant,
• $q_1$ and $q_2$ are the magnitudes of the charges,
• $r$ is the distance between the charges.

Step 2: Calculate Forces Exerted by $q_1$ and $q_2$ on $q_3$

Force exerted by $q_1$ on $q_3$:

• $q_1 = +4.00 \, \text{nC} = 4.00 \times 10^{-9} \, \text{C}$,
• $q_3 = -6.00 \, \text{nC} = -6.00 \times 10^{-9} \, \text{C}$,
• Distance between $q_1$ (at $x = 0.200 \, \text{m}$) and $q_3$ (at the origin) is $r_1 = 0.200 \, \text{m}$.

Using Coulomb's law: $F_{13} = k \frac{|q_1 q_3|}{r_1^2} = 8.99 \times 10^9 \frac{|(4.00 \times 10^{-9})(-6.00 \times 10^{-9})|}{(0.200)^2}$ $F_{13} = 8.99 \times 10^9 \times \frac{24.00 \times 10^{-18}}{0.0400} = 5.394 \times 10^{-7} \, \text{N}$ Since $q_1$ is positive and $q_3$ is negative, the force will be attractive, so $q_3$ will be pulled towards $q_1$. The force is directed to the right (positive $x$-axis).

Force exerted by $q_2$ on $q_3$:

• $q_2 = +6.00 \, \text{nC} = 6.00 \times 10^{-9} \, \text{C}$,
• Distance between $q_2$ (at $x = -0.500 \, \text{m}$) and $q_3$ (at the origin) is $r_2 = 0.500 \, \text{m}$.

Using Coulomb's law: $F_{23} = k \frac{|q_2 q_3|}{r_2^2} = 8.99 \times 10^9 \frac{|(6.00 \times 10^{-9})(-6.00 \times 10^{-9})|}{(0.500)^2}$ $F_{23} = 8.99 \times 10^9 \times \frac{36.00 \times 10^{-18}}{0.2500} = 1.295 \times 10^{-7} \, \text{N}$ Again, since $q_2$ is positive and $q_3$ is negative, the force will be attractive, so $q_3$ will be pulled towards $q_2$. The force is directed to the left (negative $x$-axis).

Step 3: Calculate the Net Force

Now, we calculate the net force by adding the forces as vectors. The force exerted by $q_1$ pulls to the right (positive $x$-axis) and the force exerted by $q_2$ pulls to the left (negative $x$-axis).

Let:

• $F_{13}$ be the force due to $q_1$ acting to the right (positive),
• $F_{23}$ be the force due to $q_2$ acting to the left (negative).

The net force is: $F_{\text{net}} = F_{13} - F_{23} = (5.394 \times 10^{-7}) - (1.295 \times 10^{-7}) = 4.099 \times 10^{-7} \, \text{N}$ The positive result indicates that the net force is directed to the right (towards $q_1$).

Step 4: Final Answer

• Magnitude of the net force: $4.099 \times 10^{-7} \, \text{N}$,
• Direction: To the right (positive $x$-axis).

Would you like further details or have any questions?

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Here are five related questions:

1. How does the force change if $q_3$ were positive instead of negative?
2. How does the distance between charges affect the magnitude of the force?
3. What happens if $q_2$ is moved closer to the origin?
4. Can you explain how the direction of force is determined in such problems?
5. How would the net force change if the charge $q_1$ were doubled?

Tip: Remember that Coulomb's law force decreases with the square of the distance between charges.