Electric Charges and Fields

Gauss's Theorem and Total Flux

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

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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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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

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Electric Dipole - Moment, Field and Torque

A conducting sphere of radius 21 cm has an electric field of \( 8 \times 10^3 \, \text{N/C} \) at 42 cm from its center.

**Dipole moment** governs torque and energy in external field. Axial field stronger than equatorial, torque maximum at θ = 90°, zero when aligned. Work done rotating dipole relates to ΔU = pE(1 - cosθ), explaining stable equilibrium at θ = 0°. E = (k q/r²) . 8 × 10³ = 9 × 10⁹ × (q/(0.42)²) . q = (8 × 10³ × 0.1764/9 × 10⁹) = 1.57 × 10⁻⁷ C . Substituting values gives 1.57 × 10⁻⁷ 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 Dipole - Moment, Field and Torque

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Electric Dipole - Moment, Field and Torque

Which property of electric charge explains why the total charge of an isolated system remains constant even when objects

**Electric dipole** consists of charges +q and -q separated by 2a, dipole moment p = q·2a, vector from negative to positive, unit C·m. In uniform field E, torque τ = p × E, magnitude τ = p E sinθ, tending to align p with E, potential energy U = -p·E = -p E cosθ. The conservation of electric charge states that the total charge in an isolated system remains constant over time. When objects are rubbed together, charge is transferred from one to another (e.g., electrons move), but no new charge is created or destroyed. This ensures the net charge of the system stays the

Ref: NCERT > Physics Book > Electric Charges and Fields > Electric Dipole - Moment, Field and Torque

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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

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Electric Dipole - Moment, Field and Torque

What allows an electric dipole to experience a net force in a non-uniform electric field but not in a uniform one?

**Dipole moment** governs torque and energy in external field. Axial field stronger than equatorial, torque maximum at θ = 90°, zero when aligned. Work done rotating dipole relates to ΔU = pE(1 - cosθ), explaining stable equilibrium at θ = 0°. In a non-uniform field, the field strength varies across the dipole, causing unequal forces on the positive and negative charges. This results in a net force, unlike in a uniform field where equal and opposite forces cancel out. Substituting values gives Field gradient, 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 Dipole - Moment, Field and Torque

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Continuous Charge Distribution

An infinite line charge has \( E = 7.2 \times 10^5 \, \text{N/C} \) at 5 cm. What is \( \lambda \)?

**Continuous distribution** uses linear density λ = dq/dl (C/m), surface σ = dq/dA, volume ρ = dq/dV. Field of infinite line with uniform λ is E = 2kλ/r = λ/(2π ε₀ r) radially outward, derived via cylindrical Gaussian surface, showing 1/r dependence. E = (2 k λ/r) . 7.2 × 10⁵ = (2 × 9 × 10⁹ × λ/0.05) . λ = (7.2 × 10⁵ × 0.05/18 × 10⁹) = 2 × 10⁻⁶ C/m . Substituting values gives 2.0 × 10⁻⁶ C/m, 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 > Continuous Charge Distribution

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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

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Electric Dipole - Moment, Field and Torque

Two charges \( +9 \, \mu\text{C} \) and \( -6 \, \mu\text{C} \) are 45 cm apart. What is the distance from \( +9 \, \mu\

**Dipole moment** governs torque and energy in external field. Axial field stronger than equatorial, torque maximum at θ = 90°, zero when aligned. Work done rotating dipole relates to ΔU = pE(1 - cosθ), explaining stable equilibrium at θ = 0°. Let x be distance from +9 μC , then 0.45 - x from -6 μC . (9 × 10⁻⁶/x²) = (6 × 10⁻⁶/(0.45 - x)²) , 9 (0.45 - x)² = 6 x² . 1.5 (0.2025 - 0.9 x + x²) = x² , 0.30375 - 1.35 x + 1.5 x² = x² . 0.5 x² - 1.35 x + 0.30375 = 0 ,

Ref: NCERT > Physics Book > Electric Charges and Fields > Electric Dipole - Moment, Field and Torque

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Electric Dipole - Moment, Field and Torque

A metallic sphere of radius 15 cm has a charge of \( 6 \, \mu\text{C} \). What is the electric field just outside its su

**Electric dipole** consists of charges +q and -q separated by 2a, dipole moment p = q·2a, vector from negative to positive, unit C·m. In uniform field E, torque τ = p × E, magnitude τ = p E sinθ, tending to align p with E, potential energy U = -p·E = -p E cosθ. For a conductor, E = (k q/r²) at surface ( r = 0.15 m ). E = 9 × 10⁹ × (6 × 10⁻⁶/(0.15)²) = 2.4 × 10⁶ N/C . Substituting values gives 2.4 × 10⁶ 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 Dipole - Moment, Field and Torque

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