#electric flux

Gauss's Theorem and Total Flux

A closed surface has a net flux of \( 9.04 \times 10^5 \, \text{Nm}^2/\text{C} \). What is the charge enclosed?

**Gauss's theorem** states total flux through closed surface equals enclosed charge divided by free-space permittivity, Φ_total = q_enc/ε₀, ε₀ = 8.854×10⁻¹² C²/(N·m²). Result independent of shape or size, depends only on net enclosed charge, enabling charge determination from flux. Φ = (q/ε₀) . q = Φ ε₀ = 9.04 × 10⁵ × 8.854 × 10⁻¹² = 8 × 10⁻⁶ C = 8 μC . Substituting values gives 8.0 μC, 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

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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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Gauss's Theorem and Total Flux

A net flux of \( 9.04 \times 10^4 \, \text{Nm}^2/\text{C} \) passes through a closed surface. What is the charge enclose

**Gauss's theorem** states total flux through closed surface equals enclosed charge divided by free-space permittivity, Φ_total = q_enc/ε₀, ε₀ = 8.854×10⁻¹² C²/(N·m²). Result independent of shape or size, depends only on net enclosed charge, enabling charge determination from flux. Φ = (q/ε₀) . q = Φ ε₀ = 9.04 × 10⁴ × 8.854 × 10⁻¹² = 8 × 10⁻⁷ C = 0.8 μC . Substituting values gives 0.8 μC, 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

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Gauss's Theorem and Total Flux

A closed surface has a net flux of \( 7.91 \times 10^5 \, \text{Nm}^2/\text{C} \). What is the charge enclosed?

**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. Φ = (q/ε₀) . q = Φ ε₀ = 7.91 × 10⁵ × 8.854 × 10⁻¹² = 7 × 10⁻⁶ C = 7 μC . Substituting values gives 7.0 μC, 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

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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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Gauss's Theorem and Total Flux

A spherical shell has a net flux of \( 1.13 \times 10^5 \, \text{Nm}^2/\text{C} \) through it. What is the charge enclos

**Gauss's theorem** states total flux through closed surface equals enclosed charge divided by free-space permittivity, Φ_total = q_enc/ε₀, ε₀ = 8.854×10⁻¹² C²/(N·m²). Result independent of shape or size, depends only on net enclosed charge, enabling charge determination from flux. Φ = (q/ε₀) . q = Φ ε₀ = 1.13 × 10⁵ × 8.854 × 10⁻¹² = 1.0 × 10⁻⁶ C = 1 μC . Substituting values gives 1.0 μC, 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

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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

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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Electric Flux

A uniform electric field \( E = 5 \times 10^3 \, \text{N/C} \) is along the y-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 = (0.4)² = 0.16 m² along y-axis. Flux: Φ = E · Δ S = 5 × 10³ × 0.16 = 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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