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What is the force on the charge located at x=8.00 cm in Figure 18.48(a) given that q=1.00μC? **Figure** **18.48** (a) Point charges located at 3.00, 8.00, and 11.0 cm along the *x*-axis. (b) Point charges located at 1.00, 5.00, 8.00, and 14.0 cm along the *x*-axis.

Let's analyze the given figure and calculate the force on the charge located at \( x = 8.00 \, \text{cm} \) in Figure 18.48(a), where \( q = 1.00 \, \mu\text{C} \).

### Step 1: Identify the Charges and Their Positions
From the figure (Figure 18.48(a)), we have the following charges:
- At \( x = 3.00 \, \text{cm} \): charge \( +q \)
- At \( x = 8.00 \, \text{cm} \): charge \( -2q \)
- At \( x = 11.0 \, \text{cm} \): charge \( +q \)

The charge \( q = 1.00 \, \mu\text{C} \).

### Step 2: Calculate the Forces Exerted on the Charge at \( x = 8.00 \, \text{cm} \)
The force between two point charges is given by Coulomb's law:

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

where:
- \( k_e = 8.99 \times 10^9 \, \text{N} \cdot \text{m}^2/\text{C}^2 \) is the Coulomb constant,
- \( q_1 \) and \( q_2 \) are the magnitudes of the charges,
- \( r \) is the distance between the charges.

#### Forces on the Charge at \( x = 8.00 \, \text{cm} \) from Other Charges
1. **Force from the charge at \( x = 3.00 \, \text{cm} \) (Charge \( +q \))**:
   - Distance \( r_1 = 8.00 \, \text{cm} - 3.00 \, \text{cm} = 5.00 \, \text{cm} = 0.0500 \, \text{m} \)
   - Force magnitude \( F_1 \):

   \[
   F_1 = k_e \frac{|(-2q)q|}{r_1^2} = 8.99 \times 10^9 \times \frac{(2 \times 10^{-6}) \times 10^{-6}}{(0.0500)^2}
   \]
   \[
   F_1 = 7.19 \times 10^{-2} \, \text{N}
   \]
   - Direction: The force will be attractive since the charges are opposite. Therefore, the force \( F_1 \) acts towards the left.

2. **Force from the charge at \( x = 11.0 \, \text{cm} \) (Charge \( +q \))**:
   - Distance \( r_2 = 11.0 \, \text{cm} - 8.00 \, \text{cm} = 3.00 \, \text{cm} = 0.0300 \, \text{m} \)
   - Force magnitude \( F_2 \):

   \[
   F_2 = k_e \frac{|(-2q)q|}{r_2^2} = 8.99 \times 10^9 \times \frac{(2 \times 10^{-6}) \times 10^{-6}}{(0.0300)^2}
   \]
   \[
   F_2 = 2.00 \times 10^{-1} \, \text{N}
   \]
   - Direction: Again, the force will be attractive. Therefore, the force \( F_2 \) acts towards the right.

### Step 3: Calculate the Net Force on the Charge at \( x = 8.00 \, \text{cm} \)
The net force on the charge at \( x = 8.00 \, \text{cm} \) is the vector sum of the forces \( F_1 \) and \( F_2 \). Since \( F_1 \) acts to the left and \( F_2 \) acts to the right:

\[
F_{\text{net}} = F_2 - F_1 = 2.00 \times 10^{-1} \, \text{N} - 7.19 \times 10^{-2} \, \text{N}
\]
\[
F_{\text{net}} = 1.28 \times 10^{-1} \, \text{N}
\]

The net force on the charge at \( x = 8.00 \, \text{cm} \) is \( \mathbf{0.128 \, \text{N}} \) directed **to the right**.

Would you like more details or have any questions?

Here are some related questions you might find interesting:
1. What happens if the charge at \( x = 11.0 \, \text{cm} \) is doubled?
2. How would the force change if all charges were negative?
3. What is the potential energy of the system of charges in this arrangement?
4. How does the net force change if the charge at \( x = 8.00 \, \text{cm} \) is reduced by half?
5. What would be the force on a charge placed at \( x = 0 \, \text{cm} \) in this configuration?

**Tip:** Remember that the direction of the force depends on whether the charges are like or unlike; opposite charges attract, and like charges repel.

Learn how to calculate the force on a charge located at x=8.00 cm using Coulomb's Law. This problem involves identifying charges at specific positions along the x-axis and applying Coulomb's Law to determine the net force on the charge. Suitable for college-level physics students.

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