To solve this problem, we will use Coulomb's Law to calculate the force between each pair of charges, and then we will determine the total force exerted on $q_3$ by $q_1$ and $q_2$.

Step 1: Coulomb's Law

The force between two point charges is given by Coulomb's Law:

$F = k_e \frac{|q_1 q_2|}{r^2}$

where:

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

Step 2: Force on $q_3$ due to $q_1$

Charge $q_1 = +3 \, \text{nC} = 3 \times 10^{-9} \, \text{C}$ is located at $x = 0.26 \, \text{m}$, and charge $q_3 = -8 \, \text{nC} = -8 \times 10^{-9} \, \text{C}$ is located at $x = 0 \, \text{m}$. The distance between $q_1$ and $q_3$ is:

$r_{13} = |0.26 - 0| = 0.26 \, \text{m}$

The magnitude of the force between $q_1$ and $q_3$ is:

$F_{13} = k_e \frac{|q_1 q_3|}{r_{13}^2} = (8.99 \times 10^9) \frac{|3 \times 10^{-9} \times (-8 \times 10^{-9})|}{(0.26)^2}$

Step 3: Force on $q_3$ due to $q_2$

Charge $q_2 = +6 \, \text{nC} = 6 \times 10^{-9} \, \text{C}$ is located at $x = -0.45 \, \text{m}$. The distance between $q_2$ and $q_3$ is:

$r_{23} = |0 - (-0.45)| = 0.45 \, \text{m}$

The magnitude of the force between $q_2$ and $q_3$ is:

$F_{23} = k_e \frac{|q_2 q_3|}{r_{23}^2} = (8.99 \times 10^9) \frac{|6 \times 10^{-9} \times (-8 \times 10^{-9})|}{(0.45)^2}$

Step 4: Direction of Forces

• The force between $q_1$ and $q_3$ is attractive (since $q_1$ is positive and $q_3$ is negative), so the force will be directed towards $q_1$, i.e., in the positive x-direction.
• The force between $q_2$ and $q_3$ is also attractive (since $q_2$ is positive and $q_3$ is negative), so the force will be directed towards $q_2$, i.e., in the negative x-direction.

Step 5: Total Force on $q_3$

To find the total force, we sum the forces considering their directions:

• $F_{13}$ is positive (towards $q_1$),
• $F_{23}$ is negative (towards $q_2$).

The total force is:

$F_{\text{total}} = F_{13} - F_{23}$

Let me compute these values.The forces between the charges are as follows:

• The magnitude of the force between $q_1$ and $q_3$ is $F_{13} = 3.19 \times 10^{-6} \, \text{N}$ in the positive x-direction.
• The magnitude of the force between $q_2$ and $q_3$ is $F_{23} = 2.13 \times 10^{-6} \, \text{N}$ in the negative x-direction.

Thus, the total force exerted on $q_3$ is:

$F_{\text{total}} = 1.06 \times 10^{-6} \, \text{N}$

This force is directed towards $q_1$ (in the positive x-direction), as $F_{13}$ is larger than $F_{23}$.

Would you like a deeper explanation or more details?

Here are some related questions:

1. What happens if $q_1$ and $q_2$ have opposite charges?
2. How does the force change if we move $q_3$ closer to $q_1$?
3. What if $q_3$ were placed exactly between $q_1$ and $q_2$?
4. How would the force change if $q_3$ had a different magnitude?
5. What if the distances between the charges were doubled?

Tip: Remember, forces between charges decrease with the square of the distance between them.