A point charge of +12μC is at a distance 6 cm vertically above the centre of a square of side 12 cm as shown in figure. The magnitude of the electric flux through the square will be __________ × 10³ Nm²/C.

Solution

Physics sir · 4 steps

The charge sits half a side above the centre, so the square is one face of a cube around it.

1. Build a cube. The side is 12 cm and the charge is 6 cm above the centre, which is exactly half a side. So the square is one face of a 12 cm cube with the charge at its centre.
2. Share the flux. By Gauss's law, the total flux through the closed cube is q/ε₀. The charge is at the centre, so all six faces sit the same way around it and each gets an equal share.

   φ_(square) = q/6ε₀ = (12 × 10⁻⁶)/(6 × 8.85 × 10⁻¹²)

3. Put in numbers. This is (2 × 10⁻⁶)/(8.85 × 10⁻¹²) = 2.26 × 10⁵ N m² C⁻¹.
4. So. The flux through the square is 226 × 10³ N m² C⁻¹, so the answer is 226.

The idea to keep. To find the flux through one flat face, picture a closed symmetric surface around the charge. If the face is one of n identical faces, its flux is q/(nε₀).

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