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14 public questions tagged with this topic.

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 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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Field Due to Special Charge Configurations

A thin spherical shell of radius 19 cm has \( q = 7 \, \mu\text{C} \). What is the electric field at 15 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 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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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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Electric Field and Electric Field Lines

A thin spherical shell of radius 31 cm has \( q = 20 \, \mu\text{C} \). What is the electric field at 35 cm from the cen

**Electric field** defined as E = F/q₀, force per unit positive test charge, unit N/C or V/m, direction along force on positive test charge. For point charge, E = k q/r² radially outward for q>0. Field lines start on positive and end on negative, density indicates strength. Outside shell: E = (k q/r²) . E = 9 × 10⁹ × (20 × 10⁻⁶/(0.35)²) = 9 × 10⁹ × (20 × 10⁻⁶/0.1225) = 1.469 × 10⁶ N/C . Substituting values gives 1.47 × 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 Field and Electric Field Lines

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Titration of Amino Acids

A peptide with a pI of 4.5 will be:

Negatively charged at pH 7 is the scientifically accurate answer to this question. Within the study of Titration of Amino Acids, this concept is well-established through extensive research and is documented in standard scientific literature. The specific properties, mechanisms, or characteristics of Negatively charged at pH 7 directly address what is being asked. Among the other options, Positively charged at pH 7, Neutral at pH 7, and Unaffected by pH changes do not correctly answer this question because they either refer to different concepts, describe properties of other molecules or processes, or represent common misconceptions about this topic.

Ref: Lehninger Principles of Biochemistry, Nelson & Cox, 8th Ed., Ch. 3

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