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#conducting shell

2 public questions tagged with this topic.

Why does the electric field inside a charged conducting shell remain unaffected by charges placed outside it?

**Gauss's law** Φ = ∮ E·dA = q_enc/ε₀ is fundamental relation between flux and enclosed charge. For charge at centre of cube, total flux = q/ε₀ distributes equally over six faces, each receiving Φ/6, but total remains q/ε₀ irrespective of cube edge. Gauss’s law shows that the field inside depends only on enclosed charge. External charges induce surface charges on the conductor, but these adjust to cancel the external field inside, leaving it zero regardless of outside charges. Substituting values gives No enclosed charge, which matches expected magnitude for this electrostatic configuration, confirming Coulomb's and Gauss's principles and charge quantization consistency.

Ref: NCERT > Physics Book > Electric Charges and Fields > Gauss's Theorem and Total Flux

Why does the electric field inside a charged conducting shell remain zero even when an external field is applied?

**Superposition principle** asserts net Coulomb force on charge equals vector sum of forces from each other charge independently, F_net = Σ F_i, where F_i = k q q_i/r_i² r̂_i. In equilateral triangle or square symmetry, components may cancel at centroid, producing equilibrium. Charges on the conductor’s surface redistribute to cancel any external field inside, creating an electrostatic shield. This shielding effect ensures the internal field is zero, as charges adjust to maintain equilibrium within the conductor. Substituting values gives Electrostatic shielding, which matches expected magnitude for this electrostatic configuration, confirming Coulomb's and Gauss's principles and charge quantization consistency.

Ref: NCERT > Physics Book > Electric Charges and Fields > Superposition Principle and Equilibrium of Charges