If a charge q is placed at the centre of a closed hemispherical non-conducting surface, the total flux passing through the flat surface would be :

Solution

Physics sir · 4 steps

The charge sits on the flat face, so complete the sphere: this closed half-sphere takes exactly half the flux.

1. Where q sits. The charge is at the centre of the flat circular face. It is on the boundary of the closed surface, so you cannot count all of q as enclosed.
2. Complete the sphere. Place an identical hemisphere below to make a full sphere around q. By Gauss's law the full sphere carries flux q/ε₀. Field lines spread out equally in all directions, so each half receives exactly half of it.

   φ = 1/2 × q/ε₀ = q/2ε₀

3. Which face carries it. The field lines from q run along the flat face, not through it. All of this flux therefore leaves through the curved part. The total flux of the closed hemispherical surface, which the options ask for, is q/2ε₀.
4. So. The total flux is q/2ε₀, which is option (b).

The idea to keep. When a charge sits on a surface, build the full symmetric shape around it. Each piece gets its share of q/ε₀: half for a flat face, a quarter for an edge, an eighth for a corner.

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