#uniform field

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

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 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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Electric Flux

A uniform field \( E = 6 \times 10^3 \, \text{N/C} \) is along the y-axis. What is the flux through a rectangle of 40 cm

**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: A = 0.4 × 0.25 = 0.1 m² . Flux: Φ = E A cos 45° = 6 × 10³ × 0.1 × (√(2)/2) = 424.26 N·m²/C . Substituting values gives 424 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 \( E = 8 \times 10^3 \, \text{N/C} \) is along the y-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 × 0.6 = 0.12 m² along y-axis. Flux: Φ = E · Δ S = 8 × 10³ × 0.12 = 960 N·m²/C . Substituting values gives 960 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 Field and Electric Field Lines

A uniform electric field \( E = 2 \times 10^3 \, \text{N/C} \) is along the z-axis. What is the flux through a rectangle

**Electric field concept** visualizes influence of source charge. Uniform field exerts constant force F = qE, and flux Φ = E·A = E A cosθ links field to area orientation, maximum when field normal to surface. Area vector Δ S = 0.3 × 0.5 = 0.15 m² along z-axis. Flux: Φ = E · Δ S = 2 × 10³ × 0.15 = 300 N·m²/C . Substituting values gives 300 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 Field and Electric Field Lines

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Electric Flux

A uniform field \( E = 4 \times 10^3 \, \text{N/C} \) is along the z-axis. What is the flux through a rectangle of 30 cm

**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 = 0.3 × 0.2 = 0.06 m² . Flux: Φ = E A cos 0° = 4 × 10³ × 0.06 = 240 N·m²/C . Substituting values gives 240 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 Field and Electric Field Lines

Why does the torque on an electric dipole in a uniform field depend on the sine of the angle between the dipole and the

**Electric field** defined as E = F/q₀, force per unit positive test charge, unit N/C or V/m, direction along force on positive test charge. For point charge, E = k q/r² radially outward for q>0. Field lines start on positive and end on negative, density indicates strength. Torque ( tau = pE sin θ ) arises from the couple formed by forces on the dipole’s charges. The perpendicular component of the field to the dipole axis determines the rotational effect, which is maximum at 90° and zero when aligned (sin 0° = 0). Substituting values gives Perpendicular component, which matches expected magnitude for this electrostatic

Ref: NCERT > Physics Book > Electric Charges and Fields > Electric Field and Electric Field Lines

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Gauss's Theorem and Total Flux

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

**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. Flux through face at x = 0 : Φ = E × A = 5 × 10³ × (0.3)² = 450 N·m²/C (inward). Flux through face at x = 0.3 : 450 N·m²/C (outward). Net flux: 450 - 450 = 0 N·m²/C (no charge enclosed). Substituting values gives 0 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 > Gauss's Theorem and Total Flux

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