Two charges are located in the x-y plane. If qı = -3.05 nC and is located at (x = 0.00 m, y = 0.960 m), and the second charge has magnitude of q2 = 4.60 nC and is located at (x = 1.40 m, y = 0.600 m), calculate the x and y components, Ex and Ey, of the electric field, É, in component form at the origin, (0,0). The Coulomb force constant is 1/(4reo) = 8.99 × 10° N · m²/C². Ex = N/C Ey = N/C
Two charges are located in the x-y plane. If qı = -3.05 nC and is located at (x = 0.00 m, y = 0.960 m), and the second charge has magnitude of q2 = 4.60 nC and is located at (x = 1.40 m, y = 0.600 m), calculate the x and y components, Ex and Ey, of the electric field, É, in component form at the origin, (0,0). The Coulomb force constant is 1/(4reo) = 8.99 × 10° N · m²/C². Ex = N/C Ey = N/C
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![# Electric Field Calculation for Two Charges in the x-y Plane
## Problem Statement
Two charges are located in the x-y plane. If \( q_1 = -3.05 \) nC and is located at \( (x = 0.00 \text{ m}, y = 0.960 \text{ m}) \), and the second charge has magnitude \( q_2 = 4.60 \) nC and is located at \( (x = 1.40 \text{ m}, y = 0.600 \text{ m}) \), calculate the x and y components, \( E_x \) and \( E_y \), of the electric field, \(\vec{E}\), in component form at the origin, \( (0, 0) \). The Coulomb force constant is \( \frac{1}{4\pi \epsilon_0} = 8.99 \times 10^9 \, \text{N} \cdot \text{m}^2/\text{C}^2 \).
## Step-by-Step Solution
1. **Identify and Plot Charges**:
- Charge \( q_1 = -3.05 \) nC located at \( (0.00 \text{ m}, 0.960 \text{ m}) \).
- Charge \( q_2 = 4.60 \) nC located at \( (1.40 \text{ m}, 0.600 \text{ m}) \).
2. **Calculate Distance from Each Charge to the Origin**:
- For \( q_1 \):
\[
r_{1} = \sqrt{(0.00 - 0.00)^2 + (0.960 - 0.00)^2} = 0.960 \text{ m}
\]
- For \( q_2 \):
\[
r_{2} = \sqrt{(1.40 - 0.00)^2 + (0.600 - 0.00)^2} = \sqrt{1.96 + 0.36} = 1.50 \text{ m}
\]
3. **Calculate the Contributions to the Electric Field**:
The electric field at the origin due to a point charge is given by:
\[
\vec{E} = \frac{q](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F37e1db01-06d0-4e20-aae1-26f96b6814c3%2F4c089ded-5f58-4d96-bfcb-ab6e1acdc405%2Fz3byoz_processed.png&w=3840&q=75)
Transcribed Image Text:# Electric Field Calculation for Two Charges in the x-y Plane
## Problem Statement
Two charges are located in the x-y plane. If \( q_1 = -3.05 \) nC and is located at \( (x = 0.00 \text{ m}, y = 0.960 \text{ m}) \), and the second charge has magnitude \( q_2 = 4.60 \) nC and is located at \( (x = 1.40 \text{ m}, y = 0.600 \text{ m}) \), calculate the x and y components, \( E_x \) and \( E_y \), of the electric field, \(\vec{E}\), in component form at the origin, \( (0, 0) \). The Coulomb force constant is \( \frac{1}{4\pi \epsilon_0} = 8.99 \times 10^9 \, \text{N} \cdot \text{m}^2/\text{C}^2 \).
## Step-by-Step Solution
1. **Identify and Plot Charges**:
- Charge \( q_1 = -3.05 \) nC located at \( (0.00 \text{ m}, 0.960 \text{ m}) \).
- Charge \( q_2 = 4.60 \) nC located at \( (1.40 \text{ m}, 0.600 \text{ m}) \).
2. **Calculate Distance from Each Charge to the Origin**:
- For \( q_1 \):
\[
r_{1} = \sqrt{(0.00 - 0.00)^2 + (0.960 - 0.00)^2} = 0.960 \text{ m}
\]
- For \( q_2 \):
\[
r_{2} = \sqrt{(1.40 - 0.00)^2 + (0.600 - 0.00)^2} = \sqrt{1.96 + 0.36} = 1.50 \text{ m}
\]
3. **Calculate the Contributions to the Electric Field**:
The electric field at the origin due to a point charge is given by:
\[
\vec{E} = \frac{q
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