A spherically symmetric charge distribution is considered with charge density varying as
ρ(r)= ρ₀(3/4-r/R) for r ≤ R ; zero for r>R
Where, r(r < R) is the distance from the centre O (as shown in figure). The electric field at point P will be:

Solution

Physics sir · 4 steps

The density changes with r, so add up the charge shell by shell before you apply Gauss's law.

1. Choose the surface. Draw a Gaussian sphere of radius r through P. By symmetry, E is radial and has the same value everywhere on it, so the flux is E · 4π r².
2. Charge enclosed. A thin shell of radius r' and thickness dr' holds charge ρ(r') 4π r'² dr'. Add these shells from the centre out to r.

   qᵢₙ = ∫₀ʳ ρ₀(3/4-r'/R)4π r'² dr' = 4πρ₀(r³/4-r⁴/4R) = πρ₀ r³(1-r/R)

3. Apply Gauss's law. E · 4π r² = qᵢₙ/ε₀ = (πρ₀ r³)/ε₀(1-r/R). Divide both sides by 4π r².
4. So. E = (ρ₀ r)/4ε₀(1-r/R), which is option (c).

The idea to keep. When ρ depends on r, the enclosed charge is ∫ ρ 4π r² dr. Never use ρ × 4/3π r³ unless the density is uniform.

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