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#sphere radius

3 public questions tagged with this topic.

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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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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Electric Field and Electric Field Lines

A conducting sphere of radius 16 cm has an electric field of \( 5 \times 10^3 \, \text{N/C} \) at 40 cm from its center.

**Electric field concept** visualizes influence of source charge. Uniform field exerts constant force F = qE, and flux Φ = E·A = E A cosθ links field to area orientation, maximum when field normal to surface. E = (k q/r²) . 5 × 10³ = 9 × 10⁹ × (q/(0.4)²) . q = (5 × 10³ × 0.16/9 × 10⁹) = 8.89 × 10⁻⁸ C . Substituting values gives 8.89 × 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 > Electric Field and Electric Field Lines

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