Questions: A tetrahedron has an equilateral triangle base with 12.0-cm-long edges and three equilateral triangle sides. The base is parallel to the ground, and a vertical uniform electric field of strength 150 N/C passes upward through the tetrahedron. What is the electric flux through the base? Express your answer in newton meters squared per coulomb.

Solution Steps

The base of the tetrahedron is an equilateral triangle with side length \( s = 12.0 \, \text{cm} = 0.12 \, \text{m} \). The area \( A \) of an equilateral triangle is given by the formula: \[ A = \frac{\sqrt{3}}{4} s^2 \] Substitute \( s = 0.12 \, \text{m} \) into the formula: \[ A = \frac{\sqrt{3}}{4} (0.12)^2 \]

Calculate the area using the values from Step 1: \[ A = \frac{\sqrt{3}}{4} \times 0.0144 \, \text{m}^2 \] \[ A \approx 0.00624 \, \text{m}^2 \]

The electric flux \( \Phi \) through a surface is given by the formula: \[ \Phi = E \cdot A \cdot \cos(\theta) \] where \( E = 150 \, \text{N/C} \) is the electric field strength, \( A \) is the area of the base, and \( \theta = 0^\circ \) because the electric field is perpendicular to the base. Thus, \(\cos(0^\circ) = 1\).

Substitute the values: \[ \Phi = 150 \, \text{N/C} \times 0.00624 \, \text{m}^2 \times 1 \] \[ \Phi = 0.936 \, \text{N} \cdot \text{m}^2/\text{C} \]

Final Answer