The volume charge density of a sphere of radius 6 m is 2 μ C cm⁻³. The number of lines of force per unit surface area coming out from the surface of the sphere is _______________ × 10¹⁰ NC⁻¹.
[Given : Permittivity of vacuum ε₀=8.85 × 10⁻¹² C² N⁻¹-m⁻² )

Solution

Physics sir · 4 steps

Lines of force per unit area is just the field strength E at the surface, so find E for a charged sphere.

1. What is asked. The number of field lines crossing unit area held perpendicular to them equals the field magnitude E. So we need E at the surface of the sphere.
2. Convert density. 1 cm³ = 10⁻⁶ m³, so ρ = (2 × 10⁻⁶ C)/(10⁻⁶ m³) = 2 C m⁻³.

   E = Q/(4πε₀ R²) = (ρ · 4/3π R³)/(4πε₀ R²) = (ρ R)/3ε₀

3. Put in numbers. E = (2 × 6)/(3 × 8.85 × 10⁻¹²) = 12/(26.55 × 10⁻¹²) = 4.52 × 10¹¹ N C⁻¹.
4. So. E = 45.2 × 10¹⁰ N C⁻¹, so the answer is 45.

The common slip: 136. Dropping the factor 3 and using E = (ρ R)/ε₀ gives 135.6 × 10¹⁰, which rounds to 136, not 45.

The idea to keep. Field line density equals field strength. For a uniformly charged sphere, the surface field is (ρ R)/3ε₀; always convert ρ to SI units first.

Answer from the NTA final answer key. Solution by PrepWiser.