#field direction

Conductors, Electrostatic Shielding and Dielectrics

Why can't the electric field inside a conductor in electrostatic equilibrium have a tangential component along its surfa

**Conductor in electrostatic equilibrium** has E=0 inside, charges reside on surface, potential constant throughout conductor. Hollow shell with no internal charge has zero field inside cavity, even if external field present, charges on outer surface screen interior, principle used in Faraday cage. In electrostatic equilibrium, the electric field inside a conductor is zero, and charges reside on the surface. If the electric field had a tangential component along the surface, it would exert a force on the free charges, causing them to move along the surface. This movement would contradict the c

Ref: NCERT > Physics Book > Electrostatic Potential and Capacitance > Conductors, Electrostatic Shielding and Dielectrics

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Electric Flux

A uniform electric field \( E = 4 \times 10^3 \, \text{N/C} \) is along the x-axis. What is the flux through a square of

**Electric flux** through surface measures field lines crossing it, Φ = E·A = E A cosθ for uniform field, unit N·m²/C. For circular area in xy-plane with field along z, θ = 0°, cosθ = 1, so Φ = E·πR² directly, maximum when field normal to surface. Area vector Δ S = (0.7)² = 0.49 m² along x-axis. Flux: Φ = E · Δ S = 4 × 10³ × 0.49 = 1960 N·m²/C . Substituting values gives 1960 N·m²/C, 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 > Electric Flux

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Electric Flux

A uniform field \( E = 5 \times 10^3 \, \text{N/C} \) is along the x-axis. What is the flux through a circle of radius 2

**Flux definition** Φ = ∮ E·dA links field to area orientation. For uniform E perpendicular to surface, Φ = E A, with A = πr² for circle. Inclination reduces flux by cosθ factor, sign indicating outward or inward crossing. Area: A = π (0.2)² = 0.1256 m² . Flux: Φ = E A cos 0° = 5 × 10³ × 0.1256 = 628 N·m²/C . Substituting values gives 628 N·m²/C, 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 > Electric Flux

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Electric Flux

A uniform electric field \( E = 2 \times 10^4 \, \text{N/C} \) is along the x-axis. What is the flux through a square of

**Electric flux** through surface measures field lines crossing it, Φ = E·A = E A cosθ for uniform field, unit N·m²/C. For circular area in xy-plane with field along z, θ = 0°, cosθ = 1, so Φ = E·πR² directly, maximum when field normal to surface. Area vector Δ S = 0.04 m² along x-axis. Flux: Φ = E · Δ S = (2 × 10⁴) × 0.04 = 800 N·m²/C . Substituting values gives 800 N·m²/C, 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 > Electric Flux

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