A cube is placed inside an electric field, E⃗ = 150y²ĵ. The side of the cube is 0.5 m and is placed in the field as shown in the given figure. The charge inside the cube is :

Solution

Physics sir · 4 steps

The field points along y, so only the top and bottom faces carry flux, and the bottom one sits at y = 0.

1. Which faces matter. E⃗ has only a ĵ component, so it runs parallel to the four side faces and gives them zero flux. Only the two shaded faces, which are perpendicular to y, contribute.
2. Bottom face. From the figure, the bottom face lies on the x-axis, so y = 0 there. Then E = 150 × 0² = 0 and its flux is zero.
3. Top face. The top face is at y = 0.5 m, where E = 150 × (0.5)² = 37.5 N C⁻¹, pointing outward. Its area is (0.5)² = 0.25 m², so the flux is 37.5 × 0.25 = 9.375 N m² C⁻¹.

   q = ε₀ φ = 8.854 × 10⁻¹² × 9.375 ≈ 8.3 × 10⁻¹¹ C

4. So. The total flux is 9.375 N m² C⁻¹, and by Gauss's law the enclosed charge is 8.3 × 10⁻¹¹ C, option (b).

The idea to keep. For a field along one axis, only the faces perpendicular to that axis carry flux. Read each face's coordinate from the figure, find E there, and use net flux = outgoing minus incoming.

Answer from the NTA final answer key. Solution by PrepWiser.