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Question

A conducting sphere of radius 18 cm has a surface charge density of \( 25 \, \mu\text{C/m}^2 \). What
is the electric field just outside its surface?

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Explanation

**Continuous distribution** uses linear density λ = dq/dl (C/m), surface σ = dq/dA, volume ρ = dq/dV. Field of infinite line with uniform λ is E = 2kλ/r = λ/(2π ε₀ r) radially outward, derived via cylindrical Gaussian surface, showing 1/r dependence. For a conductor: E = (sigma/ε₀) . E = (25 × 10⁻⁶/8.854 × 10⁻¹²) = 2.82 × 10⁶ N/C . Substituting values gives 2.82 × 10⁶ N/C, which matches expected magnitude for this electrostatic configuration, confirming Coulomb's and Gauss's principles and charge quantization consistency.

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