An electric field E⃗=(2 x î) N C⁻¹ exists in space. A cube of side 2 m is placed in the space as per figure given below. The electric flux through the cube is ______ Nm² / C.

Solution

Physics sir · 4 steps

The field points only along x, so only the two faces facing the x-axis carry any flux.

1. Locate the cube. From the figure, the cube starts at x = 2 m and has side 2 m, so its two x-faces sit at x = 2 m and x = 4 m. Each face has area 2 × 2 = 4 m².
2. Which faces count. E⃗ has only an î component. On the top, bottom, front and back faces, E⃗ runs along the surface, so their flux is zero.
3. Right and left faces. At x = 4, E = 8 N C⁻¹, pointing out of the cube: flux = 8 × 4 = +32. At x = 2, E = 4 N C⁻¹, pointing into the cube: flux = -4 × 4 = -16.

   φ = E(4) A - E(2) A = (8 - 4) × 4

4. So. The net flux through the cube is 16 N m² C⁻¹.

The common slip: 48. Adding both face fluxes as positive gives 32 + 16 = 48. On the left face the outward normal points along -î, so that flux is negative.

The idea to keep. For a field along one axis, net flux through a box is (field on the exit face minus field on the entry face) times the face area. Outward area vectors decide the signs.

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