The electric field in a region is given by E⃗ = 2/5E₀î + 3/5E₀ĵ with E₀ = 4.0 × 10³N/C. The flux of this field through a rectangular surface area 0.4 m² parallel to the Y-Z plane is __________ Nm²C⁻¹.

Solution

Physics sir · 4 steps

A surface parallel to the Y-Z plane faces the x-direction, so only the x-part of the field counts.

1. Find the normal. A flat surface parallel to the Y-Z plane has its area vector along the x-axis. So A⃗ = 0.4 î m².

   φ = E⃗·A⃗ = (2/5E₀î + 3/5E₀ĵ)·(0.4 î) = 2/5E₀ × 0.4

2. Drop the y-part. The ĵ component runs along the surface, so ĵ·î = 0 and it sends no field lines through the surface.
3. Put in numbers. 2/5 × 4.0 × 10³ = 1.6 × 10³ N C⁻¹. Multiplying by 0.4 m² gives 640 N m² C⁻¹.
4. So. The flux through the surface is 640 N m² C⁻¹.

The common slip: 960. Treating the surface as facing the y-direction picks the 3/5E₀ component and gives 960, not 640.

The idea to keep. Flux is E⃗·A⃗, and the area vector points perpendicular to the surface. A surface parallel to a plane has its normal along the axis missing from that plane's name.

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