Two uniform infinite sheets of charge density +σ and –σ intersect at right angles. Electricfield due to this configuration at any point in space is

The electric field due to a single infinite sheet of charge is given by E = σ/2ε₀, where σ is the surface charge density and ε₀ is the permittivity of free space. The direction of the electric field is always perpendicular to the sheet and away from the sheet if the charge is positive.

In this case, we have two sheets intersecting at right angles. One has a charge density of +σ and the other -σ.

1. The electric field due to the sheet with charge density +σ is E₁ = σ/2ε₀ in a direction away from this sheet.

2. The electric field due to the sheet with charge density -σ is E₂ = σ/2ε₀ in a direction towards this sheet (since the electric field direction is reversed for negative charges).

Since the sheets are at right angles to each other, the resultant electric field will be the vector sum of the two fields.

The magnitude of the resultant electric field E can be found using Pythagoras' theorem (since the two fields are perpendicular to each other):

E = √(E₁² + E₂²) = √[(σ/2ε₀)² + (σ/2ε₀)²] = σ/ε₀

The direction of the resultant field will be at an angle θ from the direction of E₁ (or E₂), where tan θ = E₂/E₁ = 1. Therefore, θ = 45°.

So, the electric field at any point in space due to this configuration is σ/ε₀ at an angle of 45° from the direction perpendicular to each sheet.