Practice question
Question
Why does the electric field due to a charged conducting sphere remain constant just outside its surface
regardless of its radius?
Explanation
**Independent action of charges** allows total force or field as vector sum. Geometry dictates distances to evaluation point, and resultant follows Σ k q_i/r_i², explaining zero field at symmetric centres for equal charges. The field just outside a conductor is sigma/ε₀ , where sigma is the surface charge density. For a sphere, sigma = Q/(4π r²) , but the field depends only on sigma at the surface, not r , due to equilibrium conditions. Substituting values gives Surface charge density, which matches expected magnitude for this electrostatic configuration, confirming Coulomb's and Gauss's principles and charge quantization consistency.
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