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#Gauss's law

21 public questions tagged with this topic.

Field Due to Special Charge Configurations

A thin spherical shell of radius 12 cm has a total charge of \( 9 \, \mu\text{C} \). What is the electric field at 15 cm

**Symmetric configurations** from Gauss's law produce characteristic fields. Uniformly charged infinite plane gives uniform field E = σ/(2ε₀) independent of distance due to planar symmetry, spherical shell acts as point charge outside E = kq/r² and zero inside, reflecting zero enclosed charge interior. Outside shell ( r > R ): E = (k q/r²) . k = 9 × 10⁹ N·m²/C² , q = 9 × 10⁻⁶ C , r = 0.15 m . E = 9 × 10⁹ × (9 × 10⁻⁶/(0.15)²) = 9 × 10⁹ × (9 × 10⁻⁶/0.0225) = 3.6 × 10⁶ N/C . Substituting values gives 3.6 × 10⁶ N/C, which

Ref: NCERT > Physics Book > Electric Charges and Fields > Field Due to Special Charge Configurations

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Field Due to Special Charge Configurations

A thin spherical shell of radius 8 cm has \( q = 4 \, \mu\text{C} \). What is the electric field at 6 cm from the center

**Field due to infinite plane and shells** illustrates symmetry power. Infinite plane's field remains constant because distant contributions balance, while spherical shell interior field cancels symmetrically, leading to E = 0 inside, E = kQ/r² outside. Inside shell ( r < R ): E = 0 (Gauss’s law). Substituting values gives 0 N/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 > Field Due to Special Charge Configurations

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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 Charge, Quantization and Conservation

What allows the electric field to be discontinuous across a charged surface, such as a conductor’s boundary?

**Charge conservation and quantization** govern rubbing processes where electrons transfer without creation. Total charge before and after remains equal, and any measured charge corresponds to n = q/e electrons, allowing counting of carriers from coulomb value. Surface charge density creates a field discontinuity. Inside a conductor, the field is zero, while just outside, it’s proportional to the surface charge density ( E = sigma/ε₀ ), as per Gauss’s law applied to a pillbox surface straddling the boundary. Substituting values gives Surface charge density, 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 Charge, Quantization and Conservation

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Gauss's Theorem and Total Flux

A closed surface has a net flux of \( 3.39 \times 10^5 \, \text{Nm}^2/\text{C} \). What is the charge enclosed?

**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. Φ = (q/ε₀) . q = Φ ε₀ = 3.39 × 10⁵ × 8.854 × 10⁻¹² = 3 × 10⁻⁶ C = 3 μC . Substituting values gives 3.0 μC, 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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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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Field Due to Special Charge Configurations

Why does the electric field inside a uniformly charged thin spherical shell vanish, regardless of the position within it

**Special distributions** like infinite line, plane, spherical shell demonstrate Gauss's law advantage. Outside shell, charge appears concentrated at centre; inside, q_enc = 0 implies vanishing field, key result for shielding. Gauss’s law explains this: for a spherical shell with uniform charge, a Gaussian surface inside encloses no charge (since all charge resides on the surface). Thus, the electric flux through the Gaussian surface is zero, implying the electric field inside is zero due to symmetry. Substituting values gives Gauss’s law, 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 > Field Due to Special Charge Configurations

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Electric Field and Electric Field Lines

Why does the electric field due to an infinite charged sheet remain constant regardless of its thickness?

**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. For an infinite sheet, Gauss’s law shows the field depends only on the surface charge density ( sigma/2ε₀ ). Thickness doesn’t affect the planar symmetry or enclosed charge per unit area, keeping the field constant. Substituting values gives Surface charge density, 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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Gauss's Theorem and Total Flux

A closed surface has a net flux of \( 2.26 \times 10^5 \, \text{Nm}^2/\text{C} \). 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 = 2.26 × 10⁵ × 8.854 × 10⁻¹² = 2 × 10⁻⁶ C = 2 μC . Substituting values gives 2.0 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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Gauss's Theorem and Total Flux

A closed surface has a net flux of \( 4.52 \times 10^5 \, \text{Nm}^2/\text{C} \). What is the charge enclosed?

**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. Φ = (q/ε₀) . q = Φ ε₀ = 4.52 × 10⁵ × 8.854 × 10⁻¹² = 4 × 10⁻⁶ C = 4 μC . Substituting values gives 4.0 μC, 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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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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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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