Applications of Gauss' Theorem > Electric Field Intensity Just Outside a Charged Conductor

Electric field intensity just outside a charged conductor is the strength of the electric field at a point infinitesimally close to (but outside) the surface of a charged conductor, arising due to the surface charge density on the conductor.

E = \[\dfrac{\sigma}{\varepsilon_0} \]

• Directed outward (away from surface) if the conductor is positively charged
• Directed inward (toward the surface) if the conductor is negatively charged
• σ = surface charge density (C/m²); ε0 = permittivity of free space

Step 1 - Setup: Consider a Gaussian "pillbox" straddling the conductor surface at point P, with cross-sectional area A.

Step 2 - Analyze Flux Contributions

• Flux through the inner face (inside conductor) = 0, since E = 0 inside a conductor
• Flux through the curved surface = 0, since E (outside) is perpendicular to the surface, hence parallel to the curved wall
• Flux through the outer face = E × A

Step 3 - Apply Gauss' Law

Step 4 - Substitute Charge Enclosed
Since q=σAq = \sigma Aq=σA:

Step 5 - Final Result

A charged conductor's surface can be modeled locally as two thin parallel sheets of charge density σ/2 each (representing the field pattern near a flat conducting surface):

• Each sheet contributes \[\dfrac{\sigma}{2\varepsilon_0}\] at any external point
• Total field outside = sum of both contributions = \[\frac {σ}{2ε_0}+\frac {σ}{2ε_0}=\frac {σ}{ε_0}\]

This confirms the same result via superposition logic.

Think of the conductor's surface like a one-way water tap: all the "water" (flux) exits from only one side (outside), since the inside is sealed off (E = 0). In contrast, an open charged sheet is like a tap with holes on both ends, releasing flow equally in both directions.