Three point charges are arranged as shown in the figure below. Find the magnitude and direction of the electric force on the particle q = 5.06 nC at the origin. (Let 12 = 0.275 m.) magnitude direction ° counterclockwise from the +x axis 6.00 nC 0.100 m -3.00 nC

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**Problem Statement:**

Three point charges are arranged as shown in the figure below. Find the magnitude and direction of the electric force on the particle \( q = 5.06 \, \text{nC} \) at the origin. (Let \( r_{12} = 0.275 \, \text{m} \).)

- **Magnitude:** \( \_\_\_\_\_\_\_\_ \) N
- **Direction:** \( \_\_\_\_\_\_\_\_ \) \(^\circ \text{ counterclockwise from the } +x \text{ axis} \)

**Figure Explanation:**

The figure shows a coordinate system with the y-axis and x-axis labeled. Three point charges are placed as follows:

1. A charge \( q \) with a value of \( 5.06 \, \text{nC} \) is placed at the origin (0, 0).
2. A charge \( 6.00 \, \text{nC} \) is placed on the positive x-axis at a distance \( r_{12} = 0.275 \, \text{m} \) from the origin.
3. A charge \( -3.00 \, \text{nC} \) is placed on the negative y-axis at a distance of \( 0.100 \, \text{m} \) from the origin.

The charges are positioned as follows:
- The \( 6.00 \, \text{nC} \) charge is located directly to the right of the origin.
- The \( -3.00 \, \text{nC} \) charge is located directly below the origin.

**Instructions:**

To solve the problem, calculate the resultant electric force on the charge \( q \) due to the other two charges. Use Coulomb's law to determine the forces, and then find the resultant vector by combining the individual force vectors. Finally, determine the magnitude and the angle of the resultant force vector, measured counterclockwise from the \( +x \)-axis.

**Helpful Calculations:**

1. Calculate the force \( F_1 \) on \( q \) due to the \( 6.00 \, \text{nC} \) charge.
2. Calculate the force \( F_2 \) on \( q \) due to the \( -3.00 \,
Transcribed Image Text:**Problem Statement:** Three point charges are arranged as shown in the figure below. Find the magnitude and direction of the electric force on the particle \( q = 5.06 \, \text{nC} \) at the origin. (Let \( r_{12} = 0.275 \, \text{m} \).) - **Magnitude:** \( \_\_\_\_\_\_\_\_ \) N - **Direction:** \( \_\_\_\_\_\_\_\_ \) \(^\circ \text{ counterclockwise from the } +x \text{ axis} \) **Figure Explanation:** The figure shows a coordinate system with the y-axis and x-axis labeled. Three point charges are placed as follows: 1. A charge \( q \) with a value of \( 5.06 \, \text{nC} \) is placed at the origin (0, 0). 2. A charge \( 6.00 \, \text{nC} \) is placed on the positive x-axis at a distance \( r_{12} = 0.275 \, \text{m} \) from the origin. 3. A charge \( -3.00 \, \text{nC} \) is placed on the negative y-axis at a distance of \( 0.100 \, \text{m} \) from the origin. The charges are positioned as follows: - The \( 6.00 \, \text{nC} \) charge is located directly to the right of the origin. - The \( -3.00 \, \text{nC} \) charge is located directly below the origin. **Instructions:** To solve the problem, calculate the resultant electric force on the charge \( q \) due to the other two charges. Use Coulomb's law to determine the forces, and then find the resultant vector by combining the individual force vectors. Finally, determine the magnitude and the angle of the resultant force vector, measured counterclockwise from the \( +x \)-axis. **Helpful Calculations:** 1. Calculate the force \( F_1 \) on \( q \) due to the \( 6.00 \, \text{nC} \) charge. 2. Calculate the force \( F_2 \) on \( q \) due to the \( -3.00 \,
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