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Field due to Infinite Plane Sheet

Field due to an infinite plane sheet with surface charge density σ. Uniform and independent of distance. Direction: normal to the sheet.
Class 11Class 12
Derivation

Symmetry argument

An infinite plane sheet with uniform surface charge density σ\sigma has planar symmetry. The field must be:

  • Perpendicular to the sheet (no component parallel to it by symmetry)
  • Equal in magnitude on both sides
  • Directed away from the sheet (for σ>0\sigma > 0)

Gaussian surface: the pillbox

Choose a small cylinder (pillbox) straddling the sheet, with each flat face of area AA parallel to the sheet, at equal distances on either side.

Flux through flat faces: E⃗∥n^\vec{E} \parallel \hat{n} on both faces, each contributing EAEA:

Φ=EA+EA=2EA\Phi = EA + EA = 2EA

Flux through curved side: E⃗⊥n^\vec{E} \perp \hat{n} (field is perpendicular to sheet, side is parallel to field), so contribution =0= 0.

Applying Gauss's law

qenc=σAq_{enc} = \sigma A 2EA=σAε02EA = \frac{\sigma A}{\varepsilon_0} E=σ2ε0\boxed{E = \frac{\sigma}{2\varepsilon_0}}

Key result

The field is uniform and independent of distance from the sheet. This is the defining feature of the field from an infinite plane — a direct consequence of the two-dimensional extent of the source.

Remember
For a conducting surface (not a sheet), the field just outside the surface is $E = \sigma/\varepsilon_0$ — twice the value for a sheet. The factor of 2 disappears because charge on a conductor resides only on one face of the surface.