Practice question
Question
A conducting sphere of radius 20 cm has an electric field of \( 3 \times 10^3 \, \text{N/C} \) at 40 cm
from its center. What is its charge?
Explanation
**Coulomb's law** gives force between point charges as F = k·|q₁q₂|/r², k = 1/(4π ε₀) = 9×10⁹ N·m²/C², directed along line joining charges. Like charges repel, opposite attract, magnitude scales with product of charges and inverse square of separation r². For a conductor, E = (k q/r²) outside. 3 × 10³ = 9 × 10⁹ × (q/(0.4)²) . q = (3 × 10³ × 0.16/9 × 10⁹) = 5.33 × 10⁻⁸ C . Substituting values gives 5.33 × 10⁻⁸ C, which matches expected magnitude for this electrostatic configuration, confirming Coulomb's and Gauss's principles and charge quantization consistency.
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