A square loop of sides a=1 m is held normally in front of a point charge q=1 C at a distance a/2. The flux of the electric field through the shaded region is 5/p × 1/ε₀ Nm²/C, where the value of p is ________ .

Solution

Physics sir · 4 steps

Build a cube around the charge: the loop is one face, so it gets one-sixth of the flux.

1. Complete the cube. The charge is a/2 from the centre of a square of side a. That makes the square one face of a cube of side a with q at its centre. By Gauss's law the cube carries q/ε₀, and by symmetry each of the 6 faces gets an equal share.

   φ_(square) = 1/6·q/ε₀

2. Cut into eighths. The lines through the centre split the square into 4 quadrants. Each quadrant's diagonal splits it into 2 triangles that are mirror images about the axis of the charge. So the face is 8 triangles, and each one gets the same flux, 1/8 of the face's share.
3. Count shaded parts. The two lower quadrants are fully shaded, which is 4 triangles. In the upper-left quadrant one triangle is shaded. That makes 5 of the 8 triangles.
4. So. φ = 5/8×1/6×q/ε₀ = 5/48×1/ε₀ with q = 1 C, so p = 48.

The common slip: 8. Taking the whole loop's flux as q/ε₀ instead of q/6ε₀ gives 5/8×1/ε₀, which means p = 8, not 48.

The idea to keep. A charge at distance a/2 in front of a square's centre sits at the centre of a cube, so the square gets q/6ε₀. Split the face only into pieces that are symmetric about the charge.

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