Question 2

**Setting up** The plane is at distance R from the axis, which is less than the inner radius √2R. Inside the inner conductor the field is zero, and outside the grounded outer cylinder the field is also zero. So only the part of the plane lying in the dielectric-filled annular region (√2R < r < 2R) contributes to the flux. The field in the annulus (line charge density λ = Q/ℓ, dielectric constant κ = 5): E = Q / (2π ε₀ κ r ℓ), radially outward. **Flux through the plane** Take a coordinate y along the plane (perpendicular to the axis), so a point on the plane is at r² = R² + y². The normal component of E is E·(R/r). The plane lies in the annulus for R < |y| < √3 R (since r ranges from √2R to 2R). Flux = ℓ ∫ E (R/r) dy = (Q / 2π ε₀ κ) ∫ R/(R² + y²) dy ∫ R/(R²+y²) dy = arctan(y/R), from y = R to √3R gives π/3 − π/4 = π/12; both sides (±y) give π/6. Equivalently, the strip subtends an angle π/6 at the axis, so it receives fraction (π/6)/(2π) = 1/12 of the total flux Q/(κε₀). Flux = (Q / (2π ε₀ · 5)) · (π/6) = Q / (60 ε₀) **Answer: (C)**

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