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#Gauss's law

55 public questions tagged with this topic.

EM Wave Characteristics - Transverse Nature and E/B Ratio

What does Gauss's Law for magnetism imply 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 magnetism, oint B · d A = 0 , implies that there are no magnetic monopoles, as the net magnetic flux through a closed surface is zero. Using c = fλ, E₀/B₀ = c, I_d = ε₀ dΦ_E/dt, and spectrum classification λ = c/f, evaluation yields No magnetic monopoles exist, illustrating EM wave transverse nature and Maxwell's displacement current concept.

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

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Energy Stored in Capacitor and Energy Density

Why does the electric field due to an infinite charged sheet remain constant with distance from the sheet?

**Energy stored in capacitor** U = ½ C V² = ½ Q V = Q²/(2C) (J), C capacitance (F), V voltage (V), Q charge (C). For 4 μF charged to 250 V, U=0.5×4×10⁻⁶×62500=0.125 J. Energy resides in electric field, energy density u = ½ ε₀ E² (J/m³), E field between plates. Using Gauss’s law for an infinite charged sheet with surface charge density sigma , a Gaussian surface (e.g., a cylinder perpendicular to the sheet) shows that the electric field E is perpendicular to the sheet and constant. The flux through the cylinder's end caps is E · 2A , and the enclosed charge is

Ref: NCERT > Physics Book > Electrostatic Potential and Capacitance > Energy Stored in Capacitor and Energy Density

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Conductors, Electrostatic Shielding and Dielectrics

Why does the electric field inside a uniformly charged spherical shell vary linearly with distance from the center if a

**Dielectric polarization** when slab inserted, bound charges appear reducing effective field, capacitance increases by factor K, potential difference for constant charge V = Q/C decreases, for constant voltage charge increases. Dielectric constant K = ε/ε₀ >1, e.g., K≈5 for glass. Without the point charge, the field inside a uniformly charged spherical shell is zero (Gauss’s law). With a point charge Q at the center, the field inside the shell is due to the point charge only ( E = (1/4 π ε₀) (Q/r²) ), which varies as (1/r²) , not linearly. The question may imply a misunderstanding; in standar

Ref: NCERT > Physics Book > Electrostatic Potential and Capacitance > Conductors, Electrostatic Shielding and Dielectrics

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Capacitor with Dielectric and Effect of Inserting Slab

In a system where a positively charged sphere is enclosed by an uncharged conducting shell, why does the outer surface o

**Effect of dielectric** is to reduce effective field due to polarization, bound surface charges opposite to free charges, net field E = E₀ - E_p = E₀/K. Capacitance C = Q/V = K ε₀ A/d for full filling, partial filling C = ε₀ A/(d - t + t/K) for slab thickness t. The positively charged sphere induces a negative charge on the inner surface of the shell and a positive charge on its outer surface to maintain zero field inside the conductor. When a negative charge is brought near the shell, it repels negative charges to the far side of the shell and attracts positive

Ref: NCERT > Physics Book > Electrostatic Potential and Capacitance > Capacitor with Dielectric and Effect of Inserting Slab

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Conductors, Electrostatic Shielding and Dielectrics

Why does the electric field just outside a charged conductor depend only on the surface charge density and not on the to

**Dielectric polarization** when slab inserted, bound charges appear reducing effective field, capacitance increases by factor K, potential difference for constant charge V = Q/C decreases, for constant voltage charge increases. Dielectric constant K = ε/ε₀ >1, e.g., K≈5 for glass. In electrostatic equilibrium, the field just outside a conductor is given by E = (sigma/ε₀) , where sigma is the local surface charge density. This relation arises from Gauss's law applied to a small Gaussian surface at the conductor's surface. The field depends on the charge enclosed per unit area ( sigma ), not th

Ref: NCERT > Physics Book > Electrostatic Potential and Capacitance > Conductors, Electrostatic Shielding and Dielectrics

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Potential Energy of System of Charges

In a system where a charged conductor is placed inside a larger uncharged hollow conductor, why does the inner conductor

**System of charges** potential energy is sum over pairs U = Σ k q_i q_j/r_ij, work required to assemble charges from infinity. For 28 μC and -14 μC, 0.28 m apart, U=9×10⁹×28×(-14)×10⁻¹²/0.28= -12.6 J, negative indicates bound system. When a charged conductor (charge Q ) is placed inside a larger uncharged hollow conductor, the electric field from the inner charge induces -Q on the inner surface of the hollow conductor to ensure the field inside the conductor's material is zero. Since the hollow conductor is initially uncharged, its total charge must remain zero, so +Q appears on its outer sur

Ref: NCERT > Physics Book > Electrostatic Potential and Capacitance > Potential Energy of System of Charges

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Conductors, Electrostatic Shielding and Dielectrics

Why does the electric field inside a uniformly charged spherical shell remain zero even if the shell is placed in an ext

**Dielectric polarization** when slab inserted, bound charges appear reducing effective field, capacitance increases by factor K, potential difference for constant charge V = Q/C decreases, for constant voltage charge increases. Dielectric constant K = ε/ε₀ >1, e.g., K≈5 for glass. For a uniformly charged spherical shell in electrostatic equilibrium, the electric field inside is zero regardless of external fields due to electrostatic shielding. Applying Gauss’s law inside the shell (no charge enclosed within the cavity), E = 0 . The external field induces charge redistribution on the shell's o

Ref: NCERT > Physics Book > Electrostatic Potential and Capacitance > Conductors, Electrostatic Shielding and Dielectrics

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Diamagnetism, Paramagnetism and Ferromagnetism

The net magnetic flux through a closed surface surrounding a bar magnet is:

**Paramagnetism** has small positive χ ≈ 10⁻³ to 10⁻⁵, weakly attracted towards stronger field, random moments align partially with B, magnetization decreases with temperature following Curie law χ ∝ 1/T. Materials have unpaired electrons with permanent moments. Gauss’s law for magnetism states that the net magnetic flux through any closed surface is zero, as magnetic field lines form closed loops with no monopoles. Substituting values gives Zero, which matches expected magnitude for this magnetic configuration, confirming dipole field dependence on m/r³ and torque relation τ = m B sinθ.

Ref: NCERT > Physics Book > Magnetism and Matter > Diamagnetism, Paramagnetism and Ferromagnetism

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Magnetic Field Lines, Bar Magnet and Dipole Moment

The fact that magnetic field lines do not start or end at any point is a direct result of:

**Magnetic dipole moment** quantifies strength and orientation of magnet, m = 2l × q_m where q_m pole strength. Field line concept visualizes B, with closed nature reflecting absence of magnetic monopoles, explaining non-intersection and continuity. Magnetic field lines form closed loops because there are no magnetic monopoles; every field line exiting a region must re-enter, ensuring continuity, as dictated by the absence of isolated magnetic charges in nature. Substituting values gives The absence of magnetic monopoles, which matches expected magnitude for this magnetic configuration, confir

Ref: NCERT > Physics Book > Magnetism and Matter > Magnetic Field Lines, Bar Magnet and Dipole Moment

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Electric Flux

A plane sheet has \( \sigma = 1.062 \times 10^{-10} \, \text{C/m}^2 \). What is the electric field near it?

**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. E = (sigma/2 ε₀) . E = (1.062 × 10⁻¹⁰/2 × 8.854 × 10⁻¹²) = 6 N/C . Substituting values gives 6.0 N/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 \( 17 \, \mu\text{C} \) 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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