High School

A square-based pyramid with dimensions 3.65 m by 3.65 m and a height of 1.69 m is placed in a uniform vertical electric field of 64.2 n/C. The pyramid encloses no charge. What is the electric flux through the pyramid?

Answer :

Final answer:

The electric flux through a square base pyramid with no enclosed charge in a uniform vertical electric field is the product of the electric field and the area of the base, considering only the base contributes to the flux because the sides are parallel to the field.

Explanation:

The electric flux (Φ) through the pyramid can be calculated using Gauss's law, which states that the electric flux is equal to the charge enclosed (Υ) divided by the permittivity of free space (ε0): Φ = Υ / ε0. Since the pyramid encloses no charge, Υ = 0, and so the flux through the bases and the sides of the pyramid will be zero as well. However, if the electric field is uniform and vertical, there would be flux through the base due to the perpendicular orientation of the base compared to the field lines. The flux through the base of the square can be calculated as the product of the electric field (E) and the area (A) of the base, with due consideration to the angle (θ) between the electric field and the normal to the surface (which is 0 degrees or cos(0) = 1 for the base).

The area (A) of the base is the square of one side, so A = 3.65 m * 3.65 m = 13.3225 m2. Thus, the electric flux is given by: Φ = E × A = 64.2 N/C × 13.3225 m2 = 855.693 Nm2/C. Since the sides of the pyramid are slanted, they contribute no net flux because the electric field is parallel to the sides, resulting in no electric field lines piercing through the sides of the pyramid.