#cube

EMF, Internal Resistance and Cells Combination

A \( 10 \, \text{V} \) battery with negligible internal resistance is connected to a cubical network of 12 resistors, ea

**Cells combination** series ε_eq = Σ ε_i, r_eq = Σ r_i, parallel for identical cells ε_eq = ε, r_eq = r/n, n number of cells. Maximum current when external R = r_eq, power transfer theorem, explaining why matching resistances maximizes power. Equivalent resistance: Rₑq = (5/6) R = (5/6) × 1.5 = 1.25 Ω . Total current: Itₒtₐl = (V/Rₑq) = (10/1.25) = 8 A . Corner current: I = (Itₒtₐl/3) = (8/3) ≈ 2.67 A . Applying I = n e A v_d, R = ρ l/A, R_t = R₀[1+αΔT], Kirchhoff's ΣI=0, ΣV=0, R_eq series/parallel, V = ε - I r and P =

Ref: NCERT > Physics Book > Current Electricity > EMF, Internal Resistance and Cells Combination

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Gauss's Theorem and Total Flux

A charge of \( 8 \, \mu\text{C} \) is at the center of a cube of edge 25 cm. What is the total flux through the cube?

**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. Total flux: Φ = (q/ε₀) . Φ = (8 × 10⁻⁶/8.854 × 10⁻¹²) = 9.03 × 10⁵ N·m²/C . Substituting values gives 9.03 × 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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Electric Flux

A charge of \( 9 \, \mu\text{C} \) is enclosed in a cube of edge 40 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) = (9 × 10⁻⁶/8.854 × 10⁻¹²) = 1.016 × 10⁶ N·m²/C . Flux per face (6 faces): Φfₐcₑ = (1.016 × 10⁶/6) = 1.693 × 10⁵ N·m²/C . Substituting values gives 1.69 × 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 charge \( 6 \, \mu\text{C} \) is at the center of a cube of edge 20 cm. What is the total electric flux through the cu

**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) . Φ = (6 × 10⁻⁶/8.854 × 10⁻¹²) = 6.78 × 10⁵ N·m²/C . Substituting values gives 6.78 × 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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