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Electricidad y magnetismo: La carga Q1, de signo positivo, tiene una magnitud de 1 micro Columb y la carga Q2, también de signo positivo, es de 1 micro columb . La distancia de q1 a q2 es de 5m, del punto cero a q1 es 10m. con el método vectorial determine la magnitud de la fuerza resultante de una carga que se encuentra entre la distancia de 10m debido a la presencia de las demás cargas 

To solve this problem using the vectorial method, we need to determine the electric forces exerted on a charge located at a specific position between the other charges.

### Problem Setup:
1. **Charge \( q_1 \)**: Located at 10 meters from the origin, with a magnitude of \(1 \mu C\) (microcoulombs).
2. **Charge \( q_2 \)**: Located 5 meters from \( q_1 \) (15 meters from the origin), with a magnitude of \(1 \mu C\).
3. **Charge \( q_0 \)**: The charge in question is placed at the origin (0 meters).

### Calculating Forces:
The force between two charges is given by Coulomb's Law:

\[
F = k_e \frac{|q_1 \cdot q_2|}{r^2}
\]

where:
- \( k_e \) is the Coulomb constant (\(8.99 \times 10^9 \, \text{N} \cdot \text{m}^2 / \text{C}^2\)),
- \( q_1 \) and \( q_2 \) are the magnitudes of the charges,
- \( r \) is the distance between the charges.

### Forces on Charge at the Origin:
1. **Force due to \( q_1 \)**:
   - Distance = 10 m
   - Force magnitude:

   \[
   F_{01} = k_e \frac{|q_0 \cdot q_1|}{(10)^2} = k_e \frac{|1 \times 10^{-6} \cdot 1 \times 10^{-6}|}{100} 
   \]

2. **Force due to \( q_2 \)**:
   - Distance = 15 m
   - Force magnitude:

   \[
   F_{02} = k_e \frac{|q_0 \cdot q_2|}{(15)^2} = k_e \frac{|1 \times 10^{-6} \cdot 1 \times 10^{-6}|}{225}
   \]

### Direction of Forces:
Both forces will act along the x-axis because the charges are aligned along this axis. Since all charges are positive, the forces will repel:

- \( F_{01} \) acts to the left (negative x-direction).
- \( F_{02} \) acts to the right (positive x-direction).

### Resultant Force:
To find the resultant force, we subtract the force due to \( q_2 \) from the force due to \( q_1 \) because they are in opposite directions:

\[
F_{R} = F_{01} - F_{02}
\]

Now let's calculate these forces.The calculated forces are as follows:

1. **Force due to \( q_1 \) (10 m away):** \( F_{01} = 8.99 \times 10^{-5} \, \text{N} \) (to the left).
2. **Force due to \( q_2 \) (15 m away):** \( F_{02} = 3.995 \times 10^{-5} \, \text{N} \) (to the right).
3. **Resultant Force \( F_R \):** \( F_R = 4.994 \times 10^{-5} \, \text{N} \) (to the left).

The resultant force acting on the charge at the origin is approximately \( 4.994 \times 10^{-5} \, \text{N} \) to the left along the x-axis.

Would you like further details or have any questions?

### Related Questions:
1. How would the forces change if one of the charges were negative?
2. What would happen to the resultant force if the distance between the charges was altered?
3. How do these forces affect the movement of the charge at the origin?
4. How would the system behave if there were more than two charges influencing the charge at the origin?
5. What is the potential energy of the system due to these charges?

**Tip:** Remember that forces between like charges are repulsive, and the magnitude of the force decreases with the square of the distance between the charges.

Electricidad y magnetismo: La carga Q1, de signo positivo, tiene una magnitud de 1 micro Columb y la carga Q2, también de signo positivo, es de 1 micro Columb. La distancia de q1 a q2 es de 5m, del punto cero a q1 es 10m. Con el método vectorial determine la magnitud de la fuerza resultante de una carga que se encuentra entre la distancia de 10m debido a la presencia de las demás cargas.

Determine la magnitud de la fuerza resultante sobre una carga colocada entre dos cargas positivas de 1 micro Coulomb cada una, usando el método vectorial y la Ley de Coulomb. Distancia entre q1 y q2 es 5 metros y q1 está ubicada a 10 metros del origen. Incluye cálculos paso a paso y la fórmula de Coulomb.

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