If h > 2R and r = 4R/5 then Φ = Q/5ε₀
If h < 8R/5 and r = 3R/5 then Φ = 0
If h > 2R and r > R then Φ = Q/ε₀
If h < 2R and r = 3R/5 then Φ = Q/5ε₀
Correct option is D. If h > 2R and r > R then Φ = Q/ε₀ (a) h > 2R, r > R Φ = Q/ε₀ clearly the Gauss' Law (b) Suppose h = 8R/5, r = 3R/5 so for h < 8R/5 Φ = 0 (c) for h = 2R, r = 4R/5 Shaded charge = 2π(1 - cos 53°) × Q/4π = Q/5 ∴ qenclosed = 2Q/5 ∴ Φ = 2Q/5ε₀ ∴ For h > 2R, r = 4R/5 ∴ Φ = 2Q/5ε₀ (d) like option C for h = 2R, r = 3R/5 qenclosed = 2 × 2π(1 - cos 37°)Q/4π = Q/5 ∴ Φ = Q/5ε₀.