As shown in figure, a cuboid lies in a region with electric field E=2 x² î-4 y ĵ+6 k̂ N / C. The magnitude of charge within the cuboid is n ∈₀ C.
The value of n is _________ (if dimension of cuboid is 1 × 2 × 3 m³ )

Solution

Physics sir · 2 steps

Go face by face: only the faces where the field is non-zero and changes across the box add net flux.

1. Set up the box. From the figure, the cuboid runs from x = 0 to 1, y = 0 to 2 and z = 0 to 3 (in m). Flux through a face is the field component along its outward normal times the face area, because each component is constant over the face it points through.

   Face pair	Field on the faces	Area	Net outward flux
   x = 1 and x = 0	Eₓ = 2 and 0	2 × 3 = 6 m²	2 × 6 - 0 = +12
   y = 2 and y = 0	E_y = -8 and 0	1 × 3 = 3 m²	-8 × 3 - 0 = -24
   z = 3 and z = 0	E_z = 6 on both	1 × 2 = 2 m²	12 - 12 = 0

   φₙₑₜ = 12 - 24 + 0 = -12 N m² C⁻¹ = qᵢₙ/ε₀

2. So. qᵢₙ = -12 ε₀ C. Its magnitude is 12 ε₀, so n = 12.

The common slip: 36. Treating the y-face flux as positive and adding magnitudes gives 12 + 24 = 36. On the y = 2 face the field points inward, so that flux is negative.

The idea to keep. For a box in a non-uniform field, pair up opposite faces. A constant component cancels on its pair; only a component that changes along its own axis leaves net flux.

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