solve this---**Extraction Content:**
**Problem Stem:**
Example 7: Three charges are arranged in an equilateral triangle configuration as shown in the diagram below.
Calculate the magnitude and direction of the force experienced by each charge.
**Diagram Description:**
* **Type:** Geometric figure with labels and a coordinate system.
* **Coordinate Axes:** Y-axis is vertical, X-axis is horizontal. The origin appears to be at the location of charge (2).
* **Shapes:** An equilateral triangle is depicted.
* **Points/Charges:**
* Charge (1): Located at the top vertex, labeled + and 7.00 μC.
* Charge (2): Located at the bottom-left vertex on the X-axis, labeled + and 2.00 μC.
* Charge (3): Located at the bottom-right vertex on the X-axis, labeled - and -4.00 μC.
* **Lines:** The sides of the triangle are drawn as solid lines connecting the charges.
* **Distances/Lengths:** One side length is labeled as 0.500 m. Since it's an equilateral triangle, all sides have this length.
* **Angles:** One internal angle of the triangle is labeled as 60.0°. Since it's an equilateral triangle, all internal angles are 60°.
* **Labels and Annotations:**
* Charges are labeled (1), (2), and (3) in blue handwritten circles.
* Handwritten force diagrams are present to the right, labeled (1), (2), and (3) corresponding to the charges.
* Diagram (1): Shows two vectors originating from a point labeled with a circle. One vector labeled F₁₂ points upwards and to the right, making an angle of 60° with a horizontal reference line. The other vector labeled F₁₃ points downwards and to the right, also making an angle of 60° with the horizontal reference line. There are also handwritten dots and arrows suggesting vector addition (parallelogram method).
* Diagram (2): Shows a point labeled with a circle. A vector labeled F₂₃ points horizontally to the right. A vector labeled F₁₂ points downwards and to the right. There is a handwritten notation "-60°" near F₁₂. A dashed line is shown, possibly representing the direction of F₂₃.
* Diagram (3): Shows a point labeled with a circle. A vector labeled F₃₂ points horizontally to the left. A vector labeled F₃₁ points upwards and to the left. An angle of 60° is indicated near F₃₁. A dashed line extending horizontally from the point is shown. Another vector labeled F₃₁ is drawn slightly offset, with an angle of 60° indicated relative to a horizontal dashed line.
**Other Relevant Text:**
μC (microcoulombs) is the unit for charge.
m (meters) is the unit for distance.
° (degrees) is the unit for angles.
**Mathematical Formulas/Chemical Equations:**
None explicitly written in standard formula format in the main problem area, although Coulomb's Law is implied for the calculation. The handwritten annotations include vector components and angles, suggesting the use of trigonometry for force calculations.
**Table Content:**
No tables are present.
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We have three charges arranged in an equilateral triangle. Charge 1 is positive 7.00 microcoulombs at the top vertex. Charge 2 is positive 2.00 microcoulombs at the bottom left. Charge 3 is negative 4.00 microcoulombs at the bottom right. Each side of the triangle is 0.500 meters long. We need to find the magnitude and direction of the force on each charge using Coulomb's law.
Now we calculate the force magnitudes using Coulomb's law. The force between charges 1 and 2 is 0.503 Newtons. The force between charges 1 and 3 is 1.007 Newtons. The force between charges 2 and 3 is 0.288 Newtons. Like charges repel each other, while opposite charges attract. We can see the force vectors acting on each charge in the diagram.
Now let's analyze the force components on charge 1. Force F12 acts at 240 degrees with magnitude 0.503 Newtons. Force F13 acts at 300 degrees with magnitude 1.007 Newtons. Breaking these into components: F12 has x-component negative 0.252 and y-component negative 0.436. F13 has x-component positive 0.504 and y-component negative 0.872. The net x-component is 0.252 Newtons, and the net y-component is negative 1.308 Newtons.
Now we calculate the net forces on charges 2 and 3. For charge 2, the net force has magnitude 0.693 Newtons at 38.9 degrees above the positive x-axis. For charge 3, the net force has magnitude 1.177 Newtons at 137.6 degrees counterclockwise from the positive x-axis. We can see all three net force vectors in the diagram, showing both their magnitudes and directions.
Here are our final results. Charge 1 experiences a net force of 1.33 Newtons at 330 degrees, or 30 degrees below the positive x-axis. Charge 2 experiences a net force of 0.693 Newtons at 38.9 degrees above the positive x-axis. Charge 3 experiences a net force of 1.177 Newtons at 137.6 degrees counterclockwise from the positive x-axis. These results were obtained using Coulomb's law to calculate individual force magnitudes, then vector addition to find the net force on each charge.