A solid sphere of radius R contains charge distributed with volume density ρ = krᵃ, where k and a are constants. If the electric field at r = R/2 is 1/8 of its value at r = R, find a.
Physics · Mixed Chapter
JEE Mains

Question

A solid sphere of radius R contains charge distributed with volume density ρ = krᵃ, where k and a are constants. If the electric field at r = R/2 is 1/8 of its value at r = R, find a.

G1 Physics · Accepted Answer

Step 1 — Given non-uniform volume charge density: ρ(r) = krᵃ. Consider a Gaussian sphere of radius r ≤ R. Step 2 — Charge enclosed: qₑₙ꜀ = ∫₀ʳ ρ(s)4πs² ds = [4πk/(a+3)]rᵃ⁺³. Step 3 — Apply Gauss’s law: E(r)·4πr² = qₑₙ꜀/ε₀. Therefore E(r) = [k/((a+3)ε₀)]rᵃ⁺¹. Step 4 — Use the supplied ratio: E(R/2)/E(R) = (1/2)ᵃ⁺¹ = 1/8 = (1/2)³. Thus a+1 = 3 ⇒ a = 2. Answer: 2.