A charge 'q' is placed at one corner of a cube as shown in figure. The flux of electrostatic field E⃗ through the shaded area is :

Solution

Physics sir · 4 steps

Put the charge at the centre of eight cubes, then split one cube's share across its far faces.

1. Share for one cube. Place seven more identical cubes around the charge so that q sits at the centre of one big cube. That big cube encloses q, so its total flux is q/ε₀. By symmetry each small cube gets one eighth of it, q/8ε₀.
2. Share for one face. Three faces of our cube (x = 0, y = 0, z = 0) pass through the charge. The field runs along them, so their flux is zero. The other three faces (x = a, y = a, z = a) sit symmetrically about the charge, so they share q/8ε₀ equally.

   φ_(one far face) = 1/3 × q/8ε₀ = q/24ε₀

3. Read the shading. The shaded area is two triangles on far faces. One is half of the top face z = a. The other is half of the side face y = a. The top triangle is cut along the diagonal that starts right above the charge, so it gets exactly half of that face's flux, q/48ε₀. The question takes the side triangle as half its face too, q/48ε₀.
4. So. Total flux through the shaded area is q/48ε₀ + q/48ε₀ = q/24ε₀, which is option (a).

The idea to keep. Charge at a cube corner: the cube gets q/8ε₀. The three faces touching the charge get zero, and each far face gets q/24ε₀. Split further only where symmetry allows it.

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