The electric field in a region is given by E⃗ = ( 3/5E₀î + 4/5E₀ĵ )N/C. The ratio of flux of reported field through the rectangular surface of area 0.2 m² (parallel to y - z plane) to that of the surface of area 0.3 m² (parallel to x - z plane) is a : b, where a = __________ [Here î, ĵ and k̂ are unit vectors along x, y and z-axes respectively.]

Solution

Physics sir · 3 steps

A surface parallel to a plane has its area vector along the axis that plane leaves out.

1. Find the normals. The y–z plane is perpendicular to the x-axis, so the first surface has area vector 0.2 î m². The x–z plane is perpendicular to the y-axis, so the second has area vector 0.3 ĵ m².

   φ = E⃗·A⃗

2. Take dot products. Only the matching component survives each dot product. φ₁ = 3/5E₀ × 0.2 = 0.12 E₀ and φ₂ = 4/5E₀ × 0.3 = 0.24 E₀.
3. So. φ₁ : φ₂ = 0.12 : 0.24 = 1 : 2, so a = 1.

The common slip: 8. Taking the field component that lies in the plane instead of the one along its normal gives 4/5× 0.2 : 3/5× 0.3 = 8 : 9, so a = 8.

The idea to keep. For a flat surface parallel to a coordinate plane, the area vector points along the missing axis. Only the field component along that axis gives any flux.

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