https://file.aiquickdraw.com/mathbot/file/1729307170645e273no62.jpg

using the image, find the x-component of the electric field at the origin, point O. 

To find the **x-component of the electric field at the origin (O)** due to the two point charges, we need to apply Coulomb's law and consider the contributions from both charges.

### Given Data:
- \( q_1 = 8.00 \, \text{nC} = 8.00 \times 10^{-9} \, \text{C} \) at \( x = +16.0 \, \text{m} \)
- \( q_2 = 6.00 \, \text{nC} = 6.00 \times 10^{-9} \, \text{C} \) at \( x = -9.00 \, \text{m} \)

Coulomb's Law gives the magnitude of the electric field due to a point charge:
\[
E = k_e \frac{|q|}{r^2}
\]
where:
- \( E \) is the electric field,
- \( k_e = 8.99 \times 10^9 \, \text{N} \cdot \text{m}^2 / \text{C}^2 \) is the Coulomb constant,
- \( q \) is the charge,
- \( r \) is the distance from the charge to the point of interest.

### Step 1: Electric Field Due to \( q_1 \)
The distance from \( q_1 \) to the origin is \( r_1 = 16.0 \, \text{m} \). Using Coulomb's law:
\[
E_1 = k_e \frac{|q_1|}{r_1^2} = 8.99 \times 10^9 \cdot \frac{8.00 \times 10^{-9}}{(16.0)^2}
\]
\[
E_1 = 2.81 \, \text{N/C}
\]
The electric field due to \( q_1 \) points **toward the negative x-direction** because the field direction is away from the positive charge at \( +16.0 \, \text{m} \).

### Step 2: Electric Field Due to \( q_2 \)
The distance from \( q_2 \) to the origin is \( r_2 = 9.00 \, \text{m} \). Using Coulomb's law:
\[
E_2 = k_e \frac{|q_2|}{r_2^2} = 8.99 \times 10^9 \cdot \frac{6.00 \times 10^{-9}}{(9.00)^2}
\]
\[
E_2 = 6.66 \, \text{N/C}
\]
The electric field due to \( q_2 \) points **toward the positive x-direction** because the field direction is toward the negative charge at \( -9.00 \, \text{m} \).

### Step 3: Combine the Electric Fields
Now we sum the x-components of the electric fields. Both fields are along the x-axis, but they point in opposite directions:
- \( E_1 \) points in the negative x-direction (so it contributes \( -2.81 \, \text{N/C} \)),
- \( E_2 \) points in the positive x-direction (so it contributes \( +6.66 \, \text{N/C} \)).

Thus, the total x-component of the electric field at the origin is:
\[
E_x = E_2 - E_1 = 6.66 \, \text{N/C} - 2.81 \, \text{N/C} = 3.85 \, \text{N/C}
\]

### Final Answer:
The x-component of the electric field at the origin is **3.85 N/C in the positive x-direction**.

Would you like any additional details or have further questions?

Here are 5 related questions to deepen your understanding:
1. How would the result change if both charges were negative?
2. What would the y-component of the electric field be at the origin?
3. How does the distance between the charges affect the electric field strength?
4. What would happen if a third charge were placed at the origin?
5. How would you calculate the force on a third charge placed at the origin?

**Tip:** Always keep track of the direction of electric field vectors, as their direction is crucial in determining the net field.

Using the image, find the x-component of the electric field at the origin, point O.

Learn how to calculate the x-component of the electric field at the origin due to two point charges: one with q1 = 8.00 nC placed 16.0 m away and another with q2 = 6.00 nC placed 9.00 m away. This step-by-step solution uses Coulomb's law to determine the net electric field.

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