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#charge distribution

18 public questions tagged with this topic.

Electric Dipole - Moment, Field and Torque

Why can’t the electric field inside a charged insulator be zero, unlike in a conductor?

**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 insulators, charges are fixed and cannot move to cancel an internal field. If charges are present inside, they generate a field that persists, as there are no free charges to redistribute and neutralize it, unlike in conductors. Substituting values gives Lack of free charges, 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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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

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Field Due to Special Charge Configurations

A thin spherical shell of radius 12 cm has \( q = 10 \, \mu\text{C} \). What is the electric field at 14 cm from the cen

**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: E = (k q/r²) . E = 9 × 10⁹ × (10 × 10⁻⁶/(0.14)²) = 9 × 10⁹ × (10 × 10⁻⁶/0.0196) = 4.59 × 10⁶ N/C . Substituting values gives 4.59 × 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 > Field Due to Special Charge Configurations

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Superposition Principle and Equilibrium of Charges

What characteristic of conductors allows charges to distribute uniformly over their surface when placed in an external e

**Vector addition of forces** underlies multi-charge analysis. Each pair contributes independent Coulomb force, resultant obtained by resolving components along axes. Equilibrium occurs when vector sum vanishes, often at symmetric points where contributions balance. In conductors, charges (free electrons) can move freely. In an external field, they redistribute until the internal field cancels the external field, achieving equilibrium. This results in charges residing only on the surface, distributed uniformly for a spherical conductor due to symmetry. Substituting values gives Mobility of charges, 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 > Superposition Principle and Equilibrium of Charges

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Field Due to Special Charge Configurations

A thin spherical shell of radius 28 cm has \( q = 8 \, \mu\text{C} \). What is the electric field at 20 cm from the cent

**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. 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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Field Due to Special Charge Configurations

A thin spherical shell of radius 10 cm has \( q = 8 \, \mu\text{C} \). What is the electric field at 15 cm from the cent

**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. Outside shell: E = (k q/r²) . E = 9 × 10⁹ × (8 × 10⁻⁶/(0.15)²) = 9 × 10⁹ × (8 × 10⁻⁶/0.0225) = 3.2 × 10⁶ N/C . Substituting values gives 3.2 × 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 > Field Due to Special Charge Configurations

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Field Due to Special Charge Configurations

A thin spherical shell of radius 17 cm has \( q = 13 \, \mu\text{C} \). What is the electric field at 20 cm from the cen

**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. Outside shell: E = (k q/r²) . E = 9 × 10⁹ × (13 × 10⁻⁶/(0.2)²) = 9 × 10⁹ × (13 × 10⁻⁶/0.04) = 2.925 × 10⁶ N/C . Substituting values gives 2.925 × 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 > Field Due to Special Charge Configurations

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Field Due to Special Charge Configurations

A thin spherical shell of radius 14 cm has a charge of \( 7 \, \mu\text{C} \). What is the electric field at a point 16

**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 = 7 × 10⁻⁶ C , r = 0.16 m . E = 9 × 10⁹ × (7 × 10⁻⁶/(0.16)²) = 9 × 10⁹ × (7 × 10⁻⁶/0.0256) = 2.46 × 10⁶ N/C . Substituting values gives 2.46 × 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 25 cm has a charge of \( 15 \, \mu\text{C} \). What is the electric field at a point 30

**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. Outside shell ( r > R ): E = (k q/r²) . k = 9 × 10⁹ N·m²/C² , q = 15 × 10⁻⁶ C , r = 0.3 m . E = 9 × 10⁹ × (15 × 10⁻⁶/(0.3)²) = 9 × 10⁹ × (15 × 10⁻⁶/0.09) = 1.5 × 10⁶ N/C . Substituting values gives 1.5 × 10⁶ N/C, which matches expected magnitude for this electrostatic configuration,

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 24 cm has a charge of \( 16 \, \mu\text{C} \). What is the electric field at a point 30

**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. Outside shell ( r > R ): E = (k q/r²) . k = 9 × 10⁹ N·m²/C² , q = 16 × 10⁻⁶ C , r = 0.3 m . E = 9 × 10⁹ × (16 × 10⁻⁶/(0.3)²) = 9 × 10⁹ × (16 × 10⁻⁶/0.09) = 1.6 × 10⁶ N/C . Substituting values gives 1.6 × 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 > Field Due to Special Charge Configurations

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Field Due to Special Charge Configurations

A thin spherical shell of radius 13 cm has a charge of \( 9 \, \mu\text{C} \). What is the electric field at a point 18

**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 = 0.18 m . E = 9 × 10⁹ × (9 × 10⁻⁶/(0.18)²) = 9 × 10⁹ × (9 × 10⁻⁶/0.0324) = 2.5 × 10⁶ N/C . Substituting values gives 2.5 × 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 6 cm has \( q = 5 \, \mu\text{C} \). What is the electric field at 8 cm from the center

**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. Outside shell: E = (k q/r²) . E = 9 × 10⁹ × (5 × 10⁻⁶/(0.08)²) = 9 × 10⁹ × (5 × 10⁻⁶/0.0064) = 7.03 × 10⁶ N/C . Substituting values gives 7.03 × 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 > Field Due to Special Charge Configurations

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