#cube geometry

Electric Flux

A charge of \( 13 \, \mu\text{C} \) is at the center of a cube of edge 50 cm. What is the flux through one face?

**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. Total flux: Φ = (q/ε₀) = (13 × 10⁻⁶/8.854 × 10⁻¹²) = 1.468 × 10⁶ N·m²/C . Flux per face (6 faces): Φfₐcₑ = (1.468 × 10⁶/6) = 2.447 × 10⁵ N·m²/C . Substituting values gives 2.45 × 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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Gauss's Theorem and Total Flux

A charge of \( 7 \, \mu\text{C} \) is at the center of a cube of edge 35 cm. What is the total flux through the cube?

**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. Total flux: Φ = (q/ε₀) . Φ = (7 × 10⁻⁶/8.854 × 10⁻¹²) = 7.91 × 10⁵ N·m²/C . Substituting values gives 7.91 × 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 > Gauss's Theorem and Total Flux

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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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Electric Flux

A charge of \( 5 \, \mu\text{C} \) is enclosed in a cube of edge 30 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) = (5 × 10⁻⁶/8.854 × 10⁻¹²) = 5.65 × 10⁵ N·m²/C . Flux per face (6 faces): Φfₐcₑ = (5.65 × 10⁵/6) = 9.42 × 10⁴ N·m²/C . Substituting values gives 9.42 × 10⁴ 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 charge of \( 3 \, \mu\text{C} \) is enclosed in a cube of edge 15 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) = (3 × 10⁻⁶/8.854 × 10⁻¹²) = 3.39 × 10⁵ N·m²/C . Flux per face (6 faces): Φfₐcₑ = (3.39 × 10⁵/6) = 5.65 × 10⁴ N·m²/C . Substituting values gives 5.65 × 10⁴ 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 charge of \( 11 \, \mu\text{C} \) is at the center of a cube of edge 45 cm. What is the total flux through the cube?

**Closed-surface flux** depends solely on net charge inside, not external charges. This principle allows flux calculation without detailed field integration and forms cornerstone for symmetric charge distributions. Total flux: Φ = (q/ε₀) . Φ = (11 × 10⁻⁶/8.854 × 10⁻¹²) = 1.242 × 10⁶ N·m²/C . Substituting values gives 1.242 × 10⁶ N·m²/C, which matches expected magnitude for this electrostatic configuration, confirming Coulomb's and Gauss's principles and charge quantization consistency. This aligns with NCERT Class 11 treatment, emphasizing conservation, symmetry and dimensional consistency useful for CBSE, NEET and CUET.

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

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