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Question

A conductor has a surface charge density of \( 4 \times 10^{-6} \, \text{C/m}^2 \). What is the
electric field just outside it? (Take \( \varepsilon_0 = 8.85 \times 10^{-12} \, \text{C}^2
\text{N}^{-1} \text{m}^{-2} \)).

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Explanation

**Relation E = -∇V** shows field points down potential gradient. For system of opposite charges close together, equipotential near midpoint between them has V≈0, but shape distorted, not spherical, reflecting superposition of potentials V = k q₁/r₁ + k q₂/r₂. E = (sigma/ε₀) = (4 × 10⁻⁶/8.85 × 10⁻¹²) ≈ 4.52 × 10⁵ N/C . Using V = kQ/r, U = k q₁q₂/r, E = -dV/dr, C = ε₀A/d, 1/C_eq series, C_eq = ΣC parallel, U = ½ C V² and common potential V = Q_total/C_total, result 4.52 × 10⁵ N/C follows, reflecting potential-capacitance relations.

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