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Just outside a large uniformly-charged conducting sheet with surface charge density σ, the magnitude of the electric field is:
Aσ / ε₀
Bσ / (2 ε₀)
Cε₀ σ
Dσ ε₀ / 2
Answer & Solution
Correct answer: A. σ / ε₀
1. For a CONDUCTOR, charge lies on the surface and E inside conductor = 0.
2. Pillbox Gaussian surface gives E · A = σ A / ε₀ ⇒ E_outside = σ / ε₀.
3. For an isolated non-conducting (insulating) sheet, both sides have E = σ / (2 ε₀).
_Source: NCERT Class 12 Physics Ch 1 §1.10 Gauss in a Conductor vs Sheet_
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