Two charged particles are attached to an x axis: Particle 1 of charge -2.00 *10^-7 C is at position x=6.00 cm and particle 2 of charge +2.00*10^-7 C is at position x=21.0 cm. Midway between the particles, what is their net electric field in unit-vector notation?

To find the net electric field at the midpoint between two charged particles, we can follow these steps:

### Step 1: Define the problem
- **Particle 1:** Charge \( q_1 = -2.00 \times 10^{-7} \, \text{C} \) located at \( x_1 = 6.00 \, \text{cm} = 0.0600 \, \text{m} \).
- **Particle 2:** Charge \( q_2 = +2.00 \times 10^{-7} \, \text{C} \) located at \( x_2 = 21.0 \, \text{cm} = 0.210 \, \text{m} \).
- **Midpoint:** Position \( x_m = \frac{0.0600 \, \text{m} + 0.210 \, \text{m}}{2} = 0.135 \, \text{m} \).

### Step 2: Calculate the distances from the midpoint to each particle
- Distance from midpoint to Particle 1: \( r_1 = x_m - x_1 = 0.135 \, \text{m} - 0.0600 \, \text{m} = 0.0750 \, \text{m} \).
- Distance from midpoint to Particle 2: \( r_2 = x_2 - x_m = 0.210 \, \text{m} - 0.135 \, \text{m} = 0.0750 \, \text{m} \).

### Step 3: Use Coulomb's Law to find the electric field due to each charge
The electric field due to a point charge \( q \) at a distance \( r \) is given by:
\[
E = \frac{k |q|}{r^2}
\]
where \( k = 8.99 \times 10^9 \, \text{N} \cdot \text{m}^2 / \text{C}^2 \).

- **Electric field due to Particle 1 (E1):**
\[
E_1 = \frac{8.99 \times 10^9 \times 2.00 \times 10^{-7}}{(0.0750)^2} = \frac{8.99 \times 10^2}{0.005625} \, \text{N/C} = 1.598 \times 10^5 \, \text{N/C}
\]
Since Particle 1 is negatively charged, the electric field it produces at the midpoint will point toward Particle 1 (to the left).

- **Electric field due to Particle 2 (E2):**
\[
E_2 = \frac{8.99 \times 10^9 \times 2.00 \times 10^{-7}}{(0.0750)^2} = 1.598 \times 10^5 \, \text{N/C}
\]
Since Particle 2 is positively charged, the electric field it produces at the midpoint will point away from Particle 2 (to the left).

### Step 4: Find the net electric field
Since both electric fields point in the same direction (to the left along the x-axis), the net electric field at the midpoint is the sum of the magnitudes of \( E_1 \) and \( E_2 \).

\[
E_{\text{net}} = E_1 + E_2 = 1.598 \times 10^5 \, \text{N/C} + 1.598 \times 10^5 \, \text{N/C} = 3.196 \times 10^5 \, \text{N/C}
\]

### Step 5: Express the net electric field in unit-vector notation
The net electric field \( \vec{E}_{\text{net}} \) is directed to the left (negative x-direction), so in unit-vector notation:
\[
\vec{E}_{\text{net}} = -3.196 \times 10^5 \, \hat{i} \, \text{N/C}
\]

### Final Answer:
The net electric field at the midpoint is \( \vec{E}_{\text{net}} = -3.196 \times 10^5 \, \hat{i} \, \text{N/C} \).

Would you like to dive deeper into any part of this problem or have any questions?

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**Related Questions:**
1. How would the net electric field change if both charges were positive?
2. What is the force experienced by a test charge placed at the midpoint?
3. How does the electric field vary as you move closer to one of the charges?
4. What would be the potential at the midpoint due to both charges?
5. How would the electric field change if the distance between the charges was doubled?

**Tip:** When solving problems involving electric fields, carefully consider the direction of the field produced by each charge to ensure correct vector addition.

Explore how to calculate the net electric field at the midpoint between two charged particles using Coulomb's Law. This detailed step-by-step solution involves vector addition and electric field concepts, suitable for students in Grades 11-12.

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