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If σ is the surface charge density and ε₀ is the permittivity of free space, then the magnitude of the electric field at the surface of a charged conductor is _____.

Asked in GSEB Board July 2022 · Conductors and electrostatic shielding

• (1) σ/(ε₀)
• (2) (2σ)/(ε₀)
• (3) σ/(2ε₀)
• (4) (5σ)/(2ε₀)

Step-by-step solution

Why the other options are wrong

• (2) No single slip gives (2σ)/(ε₀). Sanity check: the pillbox has flux only through its outer face, so EA=(σ A)/(ε₀) and E=σ/(ε₀).
• (3) This is the field of an isolated thin sheet, which sends flux out through both faces of the pillbox. Inside a conductor the field is zero, so all the flux leaves through the outer face and the field is twice as large.
• (4) No single slip gives (5σ)/(2ε₀); Gauss's law gives exactly σ/(ε₀) just outside a conductor.

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