The electric field intensity near a conducting surface having uniform surface charge density ' σ ' is given by ( ε₀ = permittivity of free space) (a) (σ)/(ε₀) and is parallel to the surface (b) (2σ)/(ε₀) and is parallel to the surface (c) (σ)/(ε₀) and is normal to the surface (d) (2σ)/(ε₀) and is normal to the surface

Answer: (c). Gauss's law with a small pillbox at the surface gives E = (σ)/(ε₀). A tangential component would move surface charges, so in equilibrium E is normal to the surface. So the answer is (c) (σ)/(ε₀), normal to the surface.

The electric field intensity near a conducting surface having uniform… | MHT CET 2025 (23 Apr, Shift 1)

Q25MHT CET 2025 · 23 APR · SHIFT 1Easy

Electric Flux & Gauss's Law

The electric field intensity near a conducting surface having uniform surface charge density ' $\sigma$ ' is given by ( $\varepsilon_0 =$ permittivity of free space)

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Answer: (c)

Gauss's law with a small pillbox at the surface gives $E = \frac{\sigma}{\varepsilon_0}$ .
A tangential component would move surface charges, so in equilibrium $\vec E$ is normal to the surface.
So the answer is (c) $\frac{\sigma}{\varepsilon_0}$ , normal to the surface.

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Electric Flux & Gauss's Law question from MHT CET 2025 (23 Apr, Shift 1)

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