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Let's solve the problem.

We have four charges arranged in a straight line with equal spacing \( d = 69 \, \text{cm} = 0.69 \, \text{m} \) between them. All charges have the same magnitude \( q = 9.4 \, \mu\text{C} = 9.4 \times 10^{-6} \, \text{C} \), but the charge \( D \) is negative while the others are positive.

We need to find the net force acting on charge \( B \). Let's denote the forces acting on \( B \) due to \( A \), \( C \), and \( D \) as \( F_{AB} \), \( F_{CB} \), and \( F_{DB} \) respectively.

### 1. Calculating Forces:

1. **Force \( F_{AB} \) due to \( A \):**
   \[
   F_{AB} = \frac{k \cdot q \cdot q}{d^2} = \frac{(8.99 \times 10^9) \cdot (9.4 \times 10^{-6})^2}{(0.69)^2}
   \]

2. **Force \( F_{CB} \) due to \( C \):**
   \[
   F_{CB} = \frac{k \cdot q \cdot q}{d^2} = \frac{(8.99 \times 10^9) \cdot (9.4 \times 10^{-6})^2}{(0.69)^2}
   \]

3. **Force \( F_{DB} \) due to \( D \):**
   \[
   F_{DB} = \frac{k \cdot q \cdot q}{(2d)^2} = \frac{(8.99 \times 10^9) \cdot (9.4 \times 10^{-6})^2}{(2 \times 0.69)^2}
   \]

### 2. Directions of Forces:
- \( F_{AB} \) and \( F_{CB} \) act to the left (since both \( A \) and \( C \) are positive charges repelling \( B \)).
- \( F_{DB} \) acts to the right (since \( D \) is a negative charge attracting \( B \)).

### 3. Net Force on \( B \):
The net force \( F_{\text{net}} \) is given by:

\[
F_{\text{net}} = F_{DB} - (F_{AB} + F_{CB})
\]

Let's compute this step-by-step.The net force acting on charge \( B \) is approximately \( -2.92 \, \text{N} \). The negative sign indicates that the force is directed to the left.

This means the correct answer is \( 2.92 \, \text{N} \) to the left, but this specific value is not listed among the provided options. It seems there might be an error in the provided answer choices or a miscalculation in the problem setup.

Would you like a more detailed breakdown or have any specific questions? 

Here are 5 related questions for further exploration:

1. How would the net force change if the charge on \( D \) were doubled?
2. What is the effect on the net force if the distance \( d \) is halved?
3. How does the force between two charges change with increasing distance?
4. What happens to the force on \( B \) if \( A \) and \( C \) were replaced with negative charges of the same magnitude?
5. How would the net force on \( B \) change if the charge \( B \) were negative?

**Tip:** Always check the units and signs carefully when calculating forces to ensure they correctly represent the physical situation.

Four charged particles are arranged in the manner as shown below. The four have the same magnitude of charge (9.4 μC) but one is negative while others are positive. Find the net force acting on charge B if d measures 69 cm.

This physics problem focuses on electrostatics, where four charges are arranged linearly with equal spacing. Learn how to calculate the net force on charge B using Coulomb's Law with equal magnitude charges (9.4 μC), where one is negative. This detailed solution walks through the process of vector addition and calculating forces between charges. Suitable for students in Grades 11-12.

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