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Gauss's Theorem and Total Flux

This category groups questions about Gauss's theorem and how to calculate total flux through surfaces. It covers the divergence theorem, applications in electrostatics, and methods for evaluating flux in various geometries.

20 questions

Why can’t electric field lines form closed loops in electrostatics, unlike magnetic field lines?

**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. Electric field lines begin at positive charges and end at negative charges (or infinity), reflecting the conservative nature of the electrostatic field. Closed loops would imply a non-conservative field, which contradicts Coulomb’s law and Gauss’s law in static conditions. Substituting values gives Conservative nature, 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

A thin spherical shell of radius 7 cm has \( q = 3 \, \mu\text{C} \). What is the electric field at 4 cm from the center

**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. 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 > Gauss's Theorem and Total Flux

What ensures that the electric field due to a uniformly charged infinite wire decreases as \( 1/r \) instead of \( 1/r^2

**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. The cylindrical symmetry of an infinite wire, when analyzed with Gauss’s law, shows the field depends on the radial distance r with a 1/r relationship. This arises because the field spreads over a cylindrical surface, not a spherical one like a point charge. Substituting values gives Cylindrical symmetry, 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

What explains why the electric field inside a charged non-conducting sphere is non-zero and varies with position?

**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. In a non-conductor, charges are fixed and distributed throughout the volume. Gauss’s law shows the field inside depends on the enclosed charge, which increases with radius, leading to a non-zero, position-dependent field. Substituting values gives Volume charge distribution, 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

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

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

Why does Gauss’s law fail to determine the electric field for a finite charged object without symmetry?

**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. Gauss’s law requires a Gaussian surface with symmetry matching the charge distribution to simplify field calculation. For finite, asymmetric objects, the field varies in complex ways, making symmetry-based simplification impossible without additional methods. Substituting values gives Lack of symmetry, 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

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

A spherical shell has a net flux of \( 1.13 \times 10^5 \, \text{Nm}^2/\text{C} \) through it. What is the charge enclos

**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 = 1.13 × 10⁵ × 8.854 × 10⁻¹² = 1.0 × 10⁻⁶ C = 1 μC . Substituting values gives 1.0 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

Why does the electric field inside a charged conducting shell remain unaffected by charges placed outside it?

**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. Gauss’s law shows that the field inside depends only on enclosed charge. External charges induce surface charges on the conductor, but these adjust to cancel the external field inside, leaving it zero regardless of outside charges. Substituting values gives No enclosed charge, 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

Why does the electric field inside a charged spherical shell remain zero even if the shell is irregular but closed?

**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. Gauss’s law applies to any closed surface: if no charge is enclosed within the shell, the net flux through a Gaussian surface inside is zero. For a conductor, charges reside on the outer surface, ensuring no field inside, regardless of shape. Substituting values gives No enclosed charge, 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

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