JEE Main 2024 · 4 April · evening · Single correct · +4 / −1
A charge q is placed at the center of one of the surface of a cube. The flux linked with the cube is:
- (a) q/(2 ε₀)
- (b) Zero
- (c) q/(4 ε₀)
- (d) q/(8 ε₀)
Answer and solution
Answer (a) q/(2 ε₀)
Solution
Physics sir · 3 steps
Put a second cube on the other side of that face: now the charge is inside, and our cube gets half.
- Complete the symmetry. The charge sits on a face, so the cube does not enclose it fully. Imagine an identical cube glued to that face. Together the two cubes form a closed box with q inside it.
- Gauss's law. The total flux out of that combined closed surface is q/ε₀. By symmetry the two cubes share it equally.
φ_(cube) = 1/2 × q/ε₀ = q/2ε₀
- So. The flux linked with the cube is q/2ε₀, which is option (a).
The idea to keep. When a charge sits on a surface, edge or corner, add identical copies around it until it is fully enclosed, then share q/ε₀ equally among the copies.
Answer from the NTA final answer key. Solution by PrepWiser.