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#conducting sphere

23 public questions tagged with this topic.

Conductors, Electrostatic Shielding and Dielectrics

A metal sphere is placed inside a uniform electric field \( E \). What is the electric field inside the sphere after it

**Electrostatic shielding** inside hollow conducting shell field zero when no charges inside, regardless of external field, because free charges redistribute on outer surface to cancel external field inside conductor, E=0 inside material in equilibrium, consequence of Gauss's law and conductor property. In electrostatic equilibrium, the electric field inside a conductor is zero. When a metal sphere is placed in a uniform electric field, charges redistribute on its surface such that the induced field inside cancels the external field. This results in a net electric field of zero inside the sphe

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

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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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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 princ

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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