Practice question
Question
A conductor has a surface charge density of \( 4.5 \times 10^{-6} \, \text{C/m}^2 \). 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 through midpoint** of +q and -q is perpendicular bisector, V=0 everywhere on it because contributions k q/r and k(-q)/r cancel. For two opposite charges, potential at midpoint zero, but field non-zero, pointing from positive to negative, illustrating vector vs scalar nature. E = (sigma/ε₀) = (4.5 × 10⁻⁶/8.85 × 10⁻¹²) ≈ 5.085 × 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 5.085 × 10⁵ N/C follows, reflecting potential-capacitance relations.
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