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19. En la Figura, la caja triangular cerrada descansa en presencia de un campo 
eléctrico horizontal de magnitud 𝑬 = 78 𝑘𝑁/𝐶, paralela a la base de la caja. Se 
sabe que 𝑎 = 10 𝑐𝑚, 𝑏 = 30 𝑐𝑚, 𝑦 𝜃 = 60º. 
A. Hallar el flujo eléctrico a través de la superficie vertical izquierda. En unidades 
de: 𝑘𝑁𝑚2/𝐶


To solve this problem, we need to calculate the electric flux through the vertical left surface of the triangular box in the presence of a horizontal electric field \(\vec{E}\).

### Given:
- Magnitude of the electric field: \( E = 78 \ \text{kN/C} \)
- Dimensions of the triangular box: \( a = 10 \ \text{cm} = 0.1 \ \text{m} \) and \( b = 30 \ \text{cm} = 0.3 \ \text{m} \)
- Angle of inclination: \( \theta = 60^\circ \)

### Approach:

The electric flux \(\Phi\) through a surface is given by the equation:
\[
\Phi = \vec{E} \cdot \vec{A} = E \cdot A \cdot \cos \theta
\]
Where:
- \( \vec{E} \) is the electric field vector.
- \( \vec{A} \) is the area vector perpendicular to the surface.

For the left vertical surface of the triangle, the area \(A\) is calculated as:
\[
A = a \times b = 0.1 \ \text{m} \times 0.3 \ \text{m} = 0.03 \ \text{m}^2
\]

Since the electric field is horizontal and the left surface is vertical, the angle between the electric field and the area vector is \( \theta = 60^\circ \).

Therefore, the electric flux through the left vertical surface is:
\[
\Phi = E \cdot A \cdot \cos \theta
\]

Let's calculate the value of the electric flux.The electric flux through the left vertical surface of the triangular box is approximately \(1170 \ \text{kNm}^2/\text{C}\).

Would you like further details or have any questions? Here are some related questions you might find interesting:

1. How would the flux change if the angle \(\theta\) were 45° instead of 60°?
2. What would happen to the flux if the magnitude of the electric field doubled?
3. How is the direction of the electric field related to the calculation of flux?
4. What is the significance of the area vector in the context of electric flux?
5. How would the flux change if the surface were not vertical but inclined?

**Tip:** Remember that electric flux depends on both the magnitude of the electric field and the orientation of the surface relative to the field.

Learn how to calculate electric flux through the left vertical surface of a triangular box in a horizontal electric field of magnitude 78 kN/C. This problem involves vector calculus, geometric surface area calculations, and applications of electric field concepts. Suitable for university-level physics students.

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