Practice question
Question
A conductor has a surface charge density of \( 5 \times 10^{-7} \, \text{C/m}^2 \). What is the
electric field just outside it? (Take \( \varepsilon_0 = 8.85 \times 10^{-12} \, \text{C}^2
\text{N}^{-1} \text{m}^{-2} \)).
Explanation
**Spherical conductor** behaves as point charge outside, potential V = k Q/R at surface, field E = k Q/r² for r>R, zero inside for rR, so E = k Q/r² =9×10⁹×6×10⁻⁸/0.16=3375 N/C. E = (sigma/ε₀) = (5 × 10⁻⁷/8.85 × 10⁻¹²) ≈ 5.65 × 10⁴ 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 5.65 × 10⁴ N/C follows, reflecting potential-capacitance relations.
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