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#conducting sphere

21 public questions tagged with this topic.

Superposition Principle and Equilibrium of Charges

A conducting sphere of radius 25 cm has an electric field of \( 5 \times 10^3 \, \text{N/C} \) at 50 cm from its center.

**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. E = (k q/r²) . 5 × 10³ = 9 × 10⁹ × (q/(0.5)²) . q = (5 × 10³ × 0.25/9 × 10⁹) = 1.389 × 10⁻⁷ C . Substituting values gives 1.39 × 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 > Superposition Principle and Equilibrium of Charges

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Superposition Principle and Equilibrium of Charges

A conducting sphere of radius 12 cm has an electric field of \( 6 \times 10^3 \, \text{N/C} \) at 24 cm from its center.

**Independent action of charges** allows total force or field as vector sum. Geometry dictates distances to evaluation point, and resultant follows Σ k q_i/r_i², explaining zero field at symmetric centres for equal charges. E = (k q/r²) . 6 × 10³ = 9 × 10⁹ × (q/(0.24)²) . q = (6 × 10³ × 0.0576/9 × 10⁹) = 3.84 × 10⁻⁸ C . Substituting values gives 3.84 × 10⁻⁸ 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 > Superposition Principle and Equilibrium of Charges

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Continuous Charge Distribution

A conducting sphere of radius 30 cm has a surface charge density of \( 50 \, \mu\text{C/m}^2 \). What is the total charg

**Line charge concept** extends point charge to infinite wire where symmetry dictates radial field proportional to λ and inversely proportional to distance r. λ = q/L for uniform case, field direction depends on sign of λ, outward for positive. Surface area: A = 4 π r² = 4 π (0.3)² = 0.36 π m² . Charge: q = sigma A = 50 × 10⁻⁶ × 0.36 π = 5.654 × 10⁻⁵ C . Substituting values gives 5.65 × 10⁻⁵ 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 > Continuous Charge Distribution

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Coulomb's Law and Force Between Point Charges

A conducting sphere of radius 20 cm has an electric field of \( 3 \times 10^3 \, \text{N/C} \) at 40 cm from its center.

**Coulomb's law** gives force between point charges as F = k·|q₁q₂|/r², k = 1/(4π ε₀) = 9×10⁹ N·m²/C², directed along line joining charges. Like charges repel, opposite attract, magnitude scales with product of charges and inverse square of separation r². For a conductor, E = (k q/r²) outside. 3 × 10³ = 9 × 10⁹ × (q/(0.4)²) . q = (3 × 10³ × 0.16/9 × 10⁹) = 5.33 × 10⁻⁸ C . Substituting values gives 5.33 × 10⁻⁸ 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 > Coulomb's Law and Force Between Point Charges

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Coulomb's Law and Force Between Point Charges

A conducting sphere of radius 15 cm has an electric field of \( 4 \times 10^3 \, \text{N/C} \) at 30 cm from its center.

**Coulomb's law** gives force between point charges as F = k·|q₁q₂|/r², k = 1/(4π ε₀) = 9×10⁹ N·m²/C², directed along line joining charges. Like charges repel, opposite attract, magnitude scales with product of charges and inverse square of separation r². E = (k q/r²) . 4 × 10³ = 9 × 10⁹ × (q/(0.3)²) . q = (4 × 10³ × 0.09/9 × 10⁹) = 4 × 10⁻⁸ C . Substituting values gives 4.0 × 10⁻⁸ 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 > Coulomb's Law and Force Between Point Charges

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Superposition Principle and Equilibrium of Charges

Why does the electric field due to a charged conducting sphere remain constant just outside its surface regardless of it

**Independent action of charges** allows total force or field as vector sum. Geometry dictates distances to evaluation point, and resultant follows Σ k q_i/r_i², explaining zero field at symmetric centres for equal charges. The field just outside a conductor is sigma/ε₀ , where sigma is the surface charge density. For a sphere, sigma = Q/(4π r²) , but the field depends only on sigma at the surface, not r , due to equilibrium conditions. Substituting values gives Surface charge density, 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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Continuous Charge Distribution

A conducting sphere of radius 18 cm has a surface charge density of \( 25 \, \mu\text{C/m}^2 \). What is the electric fi

**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. For a conductor: E = (sigma/ε₀) . E = (25 × 10⁻⁶/8.854 × 10⁻¹²) = 2.82 × 10⁶ N/C . Substituting values gives 2.82 × 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 > Continuous Charge Distribution

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Coulomb's Law and Force Between Point Charges

A conducting sphere of radius 19 cm has an electric field of \( 3 \times 10^3 \, \text{N/C} \) at 38 cm from its center.

**Inverse-square law** for charges states F ∝ 1/r² while increasing with charge product. Using k = 9×10⁹ N·m²/C², force at distance r follows F = k q₁q₂/r², forming basis for pairwise force calculation. E = (k q/r²) . 3 × 10³ = 9 × 10⁹ × (q/(0.38)²) . q = (3 × 10³ × 0.1444/9 × 10⁹) = 4.81 × 10⁻⁸ C . Substituting values gives 4.81 × 10⁻⁸ 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 > Coulomb's Law and Force Between Point Charges

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Continuous Charge Distribution

A conducting sphere of radius 29 cm has a surface charge density of \( 50 \, \mu\text{C/m}^2 \). What is the total charg

**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. Surface area: A = 4 π r² = 4 π (0.29)² = 0.335 π m² . Charge: q = sigma A = 50 × 10⁻⁶ × 0.335 π = 5.27 × 10⁻⁵ C . Substituting values gives 5.27 × 10⁻⁵ 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 > Continuous Charge Distribution

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Continuous Charge Distribution

A conducting sphere of radius 22 cm has a surface charge density of \( 40 \, \mu\text{C/m}^2 \). What is the electric fi

**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. For a conductor: E = (sigma/ε₀) . E = (40 × 10⁻⁶/8.854 × 10⁻¹²) = 4.52 × 10⁶ N/C . Substituting values gives 4.52 × 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 > Continuous Charge Distribution

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Continuous Charge Distribution

Why does the electric field outside a charged conducting sphere resemble that of a point charge located at its center?

**Charge density formulation** allows integration over extended bodies, but highly symmetric cases yield simple expressions. Infinite line gives E ∝ λ/r, unlike point charge 1/r², reflecting different geometry of source. The spherical symmetry of the charge distribution on the conductor’s surface ensures that the field outside behaves as if all charge were at the center, as per Gauss’s law. This symmetry simplifies the field to a radial, point-charge-like pattern. Substituting values gives Spherical 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 > Continuous Charge Distribution

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Continuous Charge Distribution

A conducting sphere of radius 26 cm has a surface charge density of \( 45 \, \mu\text{C/m}^2 \). What is the electric fi

**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. For a conductor: E = (sigma/ε₀) . E = (45 × 10⁻⁶/8.854 × 10⁻¹²) = 5.08 × 10⁶ N/C . Substituting values gives 5.08 × 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 > Continuous Charge Distribution

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