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Applications of Gauss' Theorem > Electric Field due to an Infinite Plane Sheet of Charge

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Estimated time: 6 minutes
CISCE: Class 12

Introduction

An infinite plane sheet of charge is an idealized, uniformly charged flat surface extending infinitely in all directions, used to model real-world situations such as charged conducting plates or capacitor plates at points close to the surface. Because the sheet is infinite, the field it produces has a special property: it does not weaken with distance, unlike the field of a point charge.

CISCE: Class 12

Derivation

Step 1 - Setup: Consider a thin, infinite sheet with uniform surface charge density σ. By symmetry, the electric field must be perpendicular to the sheet at every point.

Step 2 - Gaussian Surface: Construct a cylindrical Gaussian surface with its axis perpendicular to the sheet, and flat circular ends of area A at points P and P′, one on each side of the sheet, equidistant from it.

Step 3 - Flux through curved surface: Since E is parallel to the curved surface everywhere, the flux through the curved part of the cylinder is zero.

Step 4 - Flux through flat ends: E is perpendicular to both flat circular ends, so the total flux through them is:

ΦE​ = EA + EA = 2EA

Step 5 - Enclosed charge: The charge enclosed within the Gaussian cylinder is:

q = σA

Step 6 - Apply Gauss's Law:

ΦE = \[\frac {q}{ε_0}\]
2EA = \[\frac {σA}{ε_0}\]

Step 7 — Final Result:

E = \[\frac {σ}{2ε_0}\]

The field is independent of the distance r from the sheet — it remains constant at every point near the sheet.

Direction Rule: The field points away from the sheet if σ is positive, and toward the sheet if σ is negative.

Validity Note: This result holds strictly for points close to the sheet, treating it as effectively infinite

CISCE: Class 12

Example

A total charge of 17.7 × 10−4 C is spread uniformly over a sheet of area 200 m2. Find the electric field near the sheet.

Given: q = 17.7 × 10-4 C, A = 200 m2

Step 1 - Find σ:

σ = \[\frac{q}{A}=\frac{17.7\times10^{-4}}{200}=8.85\times10^{-6}\operatorname{C/m}^2\]

Step 2 - Apply formula:

E = \[\frac{\sigma}{2\epsilon_0}=\frac{8.85\times10^{-6}}{2\times8.85\times10^{-12}}\]

Answer:

E = 5 × 105 N/C
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