Electric Field of Charged Plane

An infinite plane sheet of charge produces a uniform electric field that is constant in magnitude everywhere and directed perpendicular to the plane. This surprising result comes from Gauss's Law.

Setup

  • Infinite plane with uniform surface charge density $\sigma$ (C/m²)
  • Want to find $E$ at distance $d$ from the plane

Derivation Using Gauss's Law

Gaussian Surface

Cylinder (pillbox) with cross-sectional area $A$, cutting through the plane. One end cap on each side.

Analysis

Through each end cap:

  • $\vec{E}$ is parallel to $\vec{A}$ (perpendicular to plane)
  • $E$ has same magnitude on both sides
  • Flux through one cap: $EA$
  • Total flux through both caps: $2EA$

Curved surface:

  • $\vec{E}$ is perpendicular to curved surface
  • Flux: $0$

Total flux: $$\Phi_E = 2EA$$

Enclosed Charge

$$Q_{enclosed} = \sigma A$$

Apply Gauss's Law

$$\Phi_E = \frac{Q_{enclosed}}{\varepsilon_0}$$

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

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

Final Result

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

Key Properties

Property Value
Direction Perpendicular to plane
Magnitude Constant — independent of distance!
Sign Points away from positive plane

Why Constant?

As you move farther from the plane:

  • Individual contributions weaken ($\propto 1/r^2$)
  • But more charge contributes (proportional to $r^2$)
  • These effects exactly cancel

Example Calculation

Problem: An infinite plane sheet produces $E = 4.5$ N/C. Find $\sigma$.

Solution: $$\sigma = 2\varepsilon_0 E = 2(8.85 \times 10^{-12})(4.5)$$

$$\sigma = 7.97 \times 10^{-11} \text{ C/m}^2$$

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