Practice question
Question
A conductor has a surface charge density of \( 1.5 \times 10^{-6} \, \text{C/m}^2 \). What is the
electric field just outside it? (Take \( \varepsilon_0 = 8.85 \times 10^{-12} \, \text{C}^2
\text{N}^{-1} \text{m}^{-2} \)).
Explanation
**Equipotential surface** is surface where V constant, no work done moving charge along it because W = q ΔV =0 when ΔV=0. Electric field E always perpendicular to equipotential surface, direction from higher to lower potential, magnitude E = -dV/dr, steeper potential gradient means stronger field. E = (sigma/ε₀) = (1.5 × 10⁻⁶/8.85 × 10⁻¹²) ≈ 1.695 × 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 1.695 × 10⁵ N/C follows, reflecting potential-capacitance relations.
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