Two point charges are placed over the x-axis as follows: charge q1=+3nC is at x=0.26m, and charge q2=+6nC is at x=-0.45m. Find the magnitude and direction of the total force exerted by these two charges over a negative point charge q3=-8nC that is at x=0m.

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.

Learn how to calculate the magnitude and direction of the total force exerted on a charge due to two other point charges using Coulomb's Law. This example involves charges q1=+3nC, q2=+6nC, and q3=-8nC at specific positions along the x-axis. Step-by-step solution provided, suitable for high school physics students.

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