Two infinite planes each with uniform surface charge density to are kept in such a way that the angle between them is 30ᵒ. The electric field in the region shown between them is given by :

Solution

Physics sir · 3 steps

Each sheet pushes σ/2ε₀ straight away from itself. Get both directions, then add.

1. Lower plane. An infinite sheet gives a field of magnitude σ/2ε₀, pointing away from a positive sheet. The region lies above the horizontal plane, so this plane gives σ/2ε₀ y.
2. Upper plane. The inclined plane makes 30° with the x-axis, so its unit normal is ±(1/2x + √3/2y). The region lies below this plane, so the field there points down and to the left: σ/2ε₀(-1/2x - √3/2y).

   E⃗ = σ/2ε₀ y + σ/2ε₀(-1/2x - √3/2y) = σ/2ε₀[(1 - √3/2)y - x/2]

3. So. The two y-components partly cancel, and only the inclined plane gives an x-component, which points along -x. This matches option (d).

The idea to keep. For several charged sheets, give each one magnitude σ/2ε₀ along its normal, pointing away from a positive sheet on the side where you are standing. Then add them as vectors.

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