Electric Flux

Electric Flux

A plane sheet has \( \sigma = 1.062 \times 10^{-10} \, \text{C/m}^2 \). What is the electric field near it?

**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. E = (sigma/2 ε₀) . E = (1.062 × 10⁻¹⁰/2 × 8.854 × 10⁻¹²) = 6 N/C . Substituting values gives 6.0 N/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 charge of \( 17 \, \mu\text{C} \) is at the center of a cube of edge 55 cm. What is the flux through one face?

**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. Total flux: Φ = (q/ε₀) = (17 × 10⁻⁶/8.854 × 10⁻¹²) = 1.92 × 10⁶ N·m²/C . Flux per face (6 faces): Φfₐcₑ = (1.92 × 10⁶/6) = 3.2 × 10⁵ N·m²/C . Substituting values gives 3.2 × 10⁵ 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 \( E = 4 \times 10^3 \, \text{N/C} \) is along the z-axis. What is the flux through a square of

**Measure of field penetration** depends on both magnitude and projected area. Understanding angle between E and normal vector is crucial, flux zero when field parallel to surface, maximum when perpendicular. Area vector Δ S = (0.15)² = 0.0225 m² along z-axis. Flux: Φ = E · Δ S = 4 × 10³ × 0.0225 = 90 N·m²/C . Substituting values gives 90 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

What causes the electric field to be stronger near a sharp point on a charged conductor compared to a flat surface?

**Measure of field penetration** depends on both magnitude and projected area. Understanding angle between E and normal vector is crucial, flux zero when field parallel to surface, maximum when perpendicular. Charge concentrates more at sharp points due to lower surface area, increasing the surface charge density. Since the field just outside a conductor is proportional to this density, the field is stronger near points than on flatter regions. Substituting values gives Charge concentration, 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 \( E = 8 \times 10^3 \, \text{N/C} \) is along the x-axis. What is the flux through a rectangle

**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 vector Δ S = 0.2 × 0.3 = 0.06 m² along x-axis. Flux: Φ = E · Δ S = 8 × 10³ × 0.06 = 480 N·m²/C . Substituting values gives 480 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 \( E = 7 \times 10^3 \, \text{N/C} \) is along the x-axis. What is the flux through a rectangle

**Measure of field penetration** depends on both magnitude and projected area. Understanding angle between E and normal vector is crucial, flux zero when field parallel to surface, maximum when perpendicular. Area vector Δ S = 0.25 × 0.4 = 0.1 m² along x-axis. Flux: Φ = E · Δ S = 7 × 10³ × 0.1 = 700 N·m²/C . Substituting values gives 700 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 \( E = 6 \times 10^3 \, \text{N/C} \) is along the z-axis. What is the flux through a square of

**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 vector Δ S = (0.5)² = 0.25 m² along z-axis. Flux: Φ = E · Δ S = 6 × 10³ × 0.25 = 1500 N·m²/C . Substituting values gives 1500 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 \( E = 4 \times 10^3 \, \text{N/C} \) 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

What property of the electric field allows it to exert a force on a charge without physical contact?

**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. The field’s ability to act at a distance is due to its nature as a force mediator. Generated by charges, it extends through space, influencing other charges via the force F = qE , without requiring direct interaction. Substituting values gives Action at a distance, 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 \( E = 9 \times 10^3 \, \text{N/C} \) is along the x-axis. What is the flux through a rectangle of 35 cm

**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: A = 0.35 × 0.45 = 0.1575 m² . Flux: Φ = E A cos 0° = 9 × 10³ × 0.1575 = 1417.5 N·m²/C . Substituting values gives 1417.5 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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