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#electric flux

44 public questions tagged with this topic.

Production of EM Waves and Hertz Experiment

A parallel plate capacitor with plate area \( A = 0.02 \, \text{m}^2 \) and separation \( d = 5 \, \text{mm} \) is conne

**Production of EM waves** requires accelerated charge, oscillating LC circuit produces changing E and B, antenna radiates when charge accelerates, frequency determined by L and C, f=1/(2π√(LC)). Hertz used spark gap with inductor and capacitor, produced ~10⁸ Hz radio waves, detected with loop, confirmed transverse nature, reflection, refraction, polarization, speed c. Displacement current i_d = ε₀ (d Φ_E/dt) . For a capacitor, i_d = i . Given i = 2 A , we have (d Φ_E/dt) = (i/ε₀) = (2/8.85 × 10⁻¹²) ≈ 2.26 × 10¹¹ Vm/s . Using c = fλ, E₀/B₀ = c, I_d = ε₀ dΦ_E/dt, and spectrum classification λ =

Ref: NCERT > Physics Book > Electromagnetic Waves > Production of EM Waves and Hertz Experiment

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Energy, Intensity and Momentum of EM Waves

A capacitor is charged such that the electric flux through a surface between the plates changes at a rate of \( 5 \times

**Momentum of EM wave** p = U/c, U energy, radiation pressure exerts force F = I A/c, small but measurable, comet tail pushed by sunlight, solar sail concept. For E₀=45 V/m, B₀=1.5×10⁻⁷ T, intensity I =0.5×3×10⁸×8.85×10⁻¹²×45²≈2.69 W/m². Displacement current i_d = ε₀ (d Φ_E/dt) . Substituting the values, i_d = (8.85 × 10⁻¹²) × (5 × 10¹⁰) = 0.4425 A . Using c = fλ, E₀/B₀ = c, I_d = ε₀ dΦ_E/dt, and spectrum classification λ = c/f, evaluation yields 0.4425 A, illustrating EM wave transverse nature and Maxwell's displacement current concept.

Ref: NCERT > Physics Book > Electromagnetic Waves > Energy, Intensity and Momentum of EM Waves

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EM Wave Characteristics - Transverse Nature and E/B Ratio

What does Gauss's Law for electricity describe in Maxwell's equations?

**Transverse nature** means E and B perpendicular to direction, e.g., wave propagating along z, E along x, B along y, Poynting vector S = E×B/μ₀ along z, energy flow direction. E and B in phase, maxima together, ratio fixed c. Gauss's Law for electricity states that the electric flux through a closed surface is proportional to the charge enclosed, given by oint E · d A = (Q/ε₀) . Using c = fλ, E₀/B₀ = c, I_d = ε₀ dΦ_E/dt, and spectrum classification λ = c/f, evaluation yields Electric flux proportional to enclosed charge, illustrating EM wave transverse nature and Maxwell's displacement curren

Ref: NCERT > Physics Book > Electromagnetic Waves > EM Wave Characteristics - Transverse Nature and E/B Ratio

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Displacement Current and Ampere-Maxwell Law

A capacitor in a circuit has a rate of change of electric flux of \( 4 \times 10^{11} \, \text{Vm/s} \). What is the dis

**Ampere-Maxwell law** ∮ B·dl = μ₀(I_c + ε₀ dΦ_E/dt) generalizes Ampere's law, displacement current arises from time-varying electric field, source of magnetic field like conduction current. For rate of change of flux 2×10¹¹ V·m/s, I_d = ε₀×2×10¹¹ =8.85×10⁻¹²×2×10¹¹=1.77 A. Displacement current i_d = ε₀ (d Φ_E/dt) . Substituting the values, i_d = (8.85 × 10⁻¹²) × (4 × 10¹¹) = 3.54 A . Using c = fλ, E₀/B₀ = c, I_d = ε₀ dΦ_E/dt, and spectrum classification λ = c/f, evaluation yields 3.54 A, illustrating EM wave transverse nature and Maxwell's displacement current concept.

Ref: NCERT > Physics Book > Electromagnetic Waves > Displacement Current and Ampere-Maxwell Law

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Displacement Current and Ampere-Maxwell Law

In a circuit with a capacitor, if the rate of change of electric flux is \( 2 \times 10^{11} \, \text{Vm/s} \), what is

**Charging capacitor** conduction current in wires equals displacement current between plates because dQ/dt = I_c = ε₀ A dE/dt = ε₀ dΦ_E/dt = I_d, preserving charge conservation, magnetic field between plates due to I_d, same as that due to conduction current. Displacement current i_d = ε₀ (d Φ_E/dt) . Substituting the values, i_d = (8.85 × 10⁻¹²) × (2 × 10¹¹) = 1.77 A . Using c = fλ, E₀/B₀ = c, I_d = ε₀ dΦ_E/dt, and spectrum classification λ = c/f, evaluation yields 1.77 A, illustrating EM wave transverse nature and Maxwell's displacement current concept.

Ref: NCERT > Physics Book > Electromagnetic Waves > Displacement Current and Ampere-Maxwell Law

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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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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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