Derive expression for electric field due to infinite charged plane using Gauss's law

Question

Using Gauss’s law, derive the expression for the electric field due to an infinite uniformly charged plane sheet with surface charge density $\sigma$ .

(NCERT Class 12, Chapter 1)

Solution — Step by Step

By symmetry, the electric field must be perpendicular to the plane and uniform in magnitude at equal distances. We choose a cylindrical Gaussian surface (pillbox) with its flat faces parallel to the charged plane, one on each side.

Let the cross-sectional area of the cylinder be $A$ .

The flux has three parts:

Curved surface: $\vec{E}$ is perpendicular to the plane, but the curved surface is parallel to $\vec{E}$ . So $\vec{E} \cdot d\vec{A} = 0$ on the curved surface.
Two flat faces: $\vec{E}$ is parallel to the outward normal on both faces. Each face contributes $EA$ .

Total flux: $\Phi = EA + EA + 0 = 2EA$

\oint \vec{E} \cdot d\vec{A} = \frac{q_{\text{enclosed}}}{\varepsilon_0}

The charge enclosed by the pillbox: $q = \sigma A$

2EA = \frac{\sigma A}{\varepsilon_0}

\boxed{E = \frac{\sigma}{2\varepsilon_0}}

The field is independent of distance from the plane.

Why This Works

Gauss’s law converts a difficult integration problem into a simple algebra problem — provided we choose the right Gaussian surface. The symmetry of an infinite plane guarantees that $E$ is the same everywhere at a given distance, allowing us to pull $E$ out of the integral.

The remarkable result: $E = \sigma/(2\varepsilon_0)$ doesn’t depend on how far you are from the plane. Whether you’re 1 cm away or 1 km away, the field is the same. This seems counterintuitive, but it’s because as you move farther away, the contribution from nearby charges decreases while the number of charges contributing increases — the two effects balance exactly.

Alternative Method — Direct integration (Coulomb’s law)

Treat the plane as a collection of ring charges. A ring of radius $r$ and width $dr$ at distance $x$ from the point contributes:

dE_x = \frac{\sigma \cdot 2\pi r\,dr}{4\pi\varepsilon_0} \cdot \frac{x}{(r^2 + x^2)^{3/2}}

Integrating from $r = 0$ to $\infty$ gives $E = \sigma/(2\varepsilon_0)$ . This is far more tedious than Gauss’s law — which is exactly why we use Gauss’s law for symmetric charge distributions.

This derivation is a 5-mark question in CBSE boards and appears almost every year. Draw a clear diagram showing the pillbox Gaussian surface, label $A$ , $E$ , and the charge density $\sigma$ . The diagram alone is worth 1-2 marks.

Common Mistake

Students often write the total flux as $EA$ instead of $2EA$ . The pillbox has two flat faces — one on each side of the plane — and the field points outward through both. Each face contributes $EA$ , so the total is $2EA$ . Missing the factor of 2 gives $E = \sigma/\varepsilon_0$ instead of $\sigma/(2\varepsilon_0)$ . Note: $\sigma/\varepsilon_0$ is the field between two parallel plates of a capacitor (where two sheets contribute), not a single sheet.

Question

Solution — Step by Step

Why This Works

Alternative Method — Direct integration (Coulomb’s law)

Common Mistake

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