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#spherical shell

43 public questions tagged with this topic.

Conductors, Electrostatic Shielding and Dielectrics

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

**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. Electrostatic Potential and Capacitance typically assumes no charge inside a spherical shell, leading to E = 0 . However, if misinterpreted as a charged dielectric sphere (common in advanced contexts but not in the PDF), the field varies as E ∝ r . Since the PDF context implies an empty shell or uniform shell charge, E = 0 . Assuming a misinterpretation, the correct context

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

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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 standard electrostatics (per the PDF), the shell

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

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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 outer surface, but this does not affect the interior, as the induced charges ensure the internal field

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

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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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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 10 cm from the cente

**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 × (4 × 10⁻⁶/(0.1)²) = 9 × 10⁹ × (4 × 10⁻⁶/0.01) = 3.6 × 10⁶ N/C . Substituting values gives 3.6 × 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 10 cm has a charge of \( 8 \, \mu\text{C} \). What is the electric field at a point 5 c

**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. Inside a thin spherical shell ( r < R ), E = 0 (Gauss’s law). Here, r = 5 cm < R = 10 cm , so E = 0 N/C . 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.

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 9 cm has \( q = 6 \, \mu\text{C} \). What is the electric field at 12 cm from the cente

**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 × (6 × 10⁻⁶/(0.12)²) = 9 × 10⁹ × (6 × 10⁻⁶/0.0144) = 3.75 × 10⁶ N/C . Substituting values gives 3.75 × 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 16 cm has a charge of \( 11 \, \mu\text{C} \). What is the electric field at a point 20

**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 = 11 × 10⁻⁶ C , r = 0.2 m . E = 9 × 10⁹ × (11 × 10⁻⁶/(0.2)²) = 9 × 10⁹ × (11 × 10⁻⁶/0.04) = 2.475 × 10⁶ N/C . Substituting values gives 2.475 × 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 15 cm has \( q = 6 \, \mu\text{C} \). What is the electric field at 10 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. 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 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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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

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Field Due to Special Charge Configurations

A thin spherical shell of radius 18 cm has \( q = 12 \, \mu\text{C} \). What is the electric field at 22 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 × (12 × 10⁻⁶/(0.22)²) = 9 × 10⁹ × (12 × 10⁻⁶/0.0484) = 2.23 × 10⁶ N/C . Substituting values gives 2.23 × 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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