Two point charges are placed as follows: charge $$ q 1=-2.50 \mathrm{nC} $$ is at $$ x=46.00 \mathrm{~m} $$ and charge $$ \mathrm{q} 2=44.80 \mathrm{nC} $$ is at the onigin. What is the total force (magnitude and direction) exerted by these two charges on a negative point charge q -3.00 nc located at $$ x=4.00 \mathrm{~m} $$ ?

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To determine the total force exerted on the negative point charge $$ q_3 = -3.00 \, \text{nC} $$ located at $$ x = 4.00 \, \text{m} $$, we need to consider the forces exerted by both $$ q_1 $$ and $$ q_2 $$.

### **Given:**
- $$ q_1 = -2.50 \, \text{nC} $$ at $$ x = 46.00 \, \text{m} $$
- $$ q_2 = 44.80 \, \text{nC} $$ at the origin ($$ x = 0 \, \text{m} $$)
- $$ q_3 = -3.00 \, \text{nC} $$ at $$ x = 4.00 \, \text{m} $$
- Coulomb's constant, $$ k = 8.988 \times 10^9 \, \text{N·m}²/\text{C}² $$

### **Step 1: Calculate the Force from $$ q_2 $$ on $$ q_3 $$ ($$ F_2 $$)**

1. **Distance between $$ q_2 $$ and $$ q_3 $$:**
   $$
   r_{23} = |4.00 \, \text{m} - 0 \, \text{m}| = 4.00 \, \text{m}
   $$
2. **Magnitude of $$ F_2 $$:**
   $$
   |F_2| = k \frac{|q_2 \cdot q_3|}{r_{23}^2} = 8.988 \times 10^9 \times \frac{44.80 \times 10^{-9} \times 3.00 \times 10^{-9}}{16} \approx 7.59 \times 10^{-8} \, \text{N}
   $$
3. **Direction of $$ F_2 $$:**
   - $$ q_2 $$ is positive and $$ q_3 $$ is negative, so the force is **attractive**.
   - Direction: **toward the origin** (left along the x-axis).

### **Step 2: Calculate the Force from $$ q_1 $$ on $$ q_3 $$ ($$ F_1 $$)**

1. **Distance between $$ q_1 $$ and $$ q_3 $$:**
   $$
   r_{13} = |46.00 \, \text{m} - 4.00 \, \text{m}| = 42.00 \, \text{m}
   $$
2. **Magnitude of $$ F_1 $$:**
   $$
   |F_1| = k \frac{|q_1 \cdot q_3|}{r_{13}^2} = 8.988 \times 10^9 \times \frac{2.50 \times 10^{-9} \times 3.00 \times 10^{-9}}{1764} \approx 3.83 \times 10^{-11} \, \text{N}
   $$
3. **Direction of $$ F_1 $$:**
   - Both $$ q_1 $$ and $$ q_3 $$ are negative, so the force is **repulsive**.
   - Direction: **away from $$ q_1 $$**, which is also **left along the x-axis**.

### **Step 3: Calculate the Total Force on $$ q_3 $$**

Both forces $$ F_1 $$ and $$ F_2 $$ act in the same direction (left along the x-axis). Therefore, the total force $$ F_{\text{total}} $$ is the sum of the magnitudes:

$$
F_{\text{total}} = F_2 + F_1 \approx 7.59 \times 10^{-8} \, \text{N} + 3.83 \times 10^{-11} \, \text{N} \approx 7.59 \times 10^{-8} \, \text{N}
$$

*(Note: $$ F_1 $$ is much smaller than $$ F_2 $$ and can be considered negligible for practical purposes.)*

### **Conclusion**

The total force exerted on the charge $$ q_3 $$ is approximately **$$ 7.6 \times 10^{-8} \, \text{N} $$** directed **to the left** along the x-axis.

**Final Answer:**

A total force of approximately 7.6 × 10⁻⁸ N acts to the left on q₃ along the x‑axis.
The total force on the negative charge $$ q_3 $$ is approximately $$ 7.6 \times 10^{-8} \, \text{N} $$ directed to the left along the x-axis.

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