Skip to content

Physics · Ch 1 — Electric Charges and Fields

Application of Gauss's Theorem: Field Due to a Uniformly Charged Infinite Plane Sheet

1.16

Application of Gauss's Theorem: Field Due to a Uniformly Charged Infinite Plane Sheet

The problem. Find the electric field near an infinite, thin, flat sheet carrying a uniform surface charge density σ\sigma (Section 1.5).

Exploiting the symmetry. An infinite plane sheet looks identical from every point along its own surface, and it possesses an obvious reflection symmetry: the field at a given perpendicular distance on one side of the sheet must be the mirror image of the field at the same perpendicular distance on the other side. The only field direction consistent with this planar symmetry is one that points exactly perpendicular to the sheet, away from it on both sides (for positive σ\sigma), with a magnitude that depends only on the perpendicular distance from the sheet -- not on position along the sheet, nor (perhaps surprisingly) on the distance from the sheet at all, as the result below will confirm.

Choosing the Gaussian surface. The symmetry is matched by choosing a small "pillbox" -- a short cylinder oriented perpendicular to the sheet, straddling it symmetrically so that it pokes out an equal perpendicular distance on each side, with its two flat circular end caps (each of area AA) parallel to the sheet, as shown in the figure.

Evaluating the flux, part by part. On each of the two flat end caps, E⃗\vec{E} is perpendicular to the sheet and hence exactly parallel to that end cap's own outward normal, with the same magnitude EE on both caps (by the reflection symmetry), so each cap contributes flux EAEA; the two caps together give

Φcaps=EA+EA=2EA\Phi_{\text{caps}} = EA+EA = 2EA

On the curved side wall of the pillbox (perpendicular to the sheet, running parallel to E⃗\vec{E}), the field is exactly parallel to this surface and perpendicular to its normal, so it contributes zero flux: Φside=0\Phi_{\text{side}}=0.

Applying Gauss's theorem. The total flux is Φ=2EA\Phi=2EA. The charge enclosed within the pillbox is just the portion of the sheet lying inside it, of area AA, carrying charge qenc=σAq_{\text{enc}}=\sigma A. Gauss's theorem then gives

2EA=σAϵ02EA = \frac{\sigma A}{\epsilon_0}

E=σ2ϵ0\boxed{E = \frac{\sigma}{2\epsilon_0}}

directed away from the sheet on both sides for a positive σ\sigma (toward the sheet on both sides for a negative σ\sigma). …

Figure 1Pillbox Gaussian surface straddling a charged plane sheet

What this figure shows. A large flat sheet is drawn stretching horizontally across the full width of the figure, shaded with a uniform pattern of plus signs to represent a uniform positive surface charge density σ\sigma. A small cylindrical pillbox is drawn straddling the sheet symmetrically, so that it pokes out by an equal small distance on both sides of the sheet, with its two flat circular end-cap faces (each of area AA) parallel to the sheet, one above and one below. On the upper end cap, an outward arrow labelled EE points straight up, away from the sheet; on the lower end cap, an equal-length outward arrow labelled EE points straight down, also away from the sheet -- illustrating that the field points away from a positively charged sheet on both sides, with the same magnitude at equal distances. The curved side wall of the pillbox (perpendicular to the sheet) is sha …