A point charge q is placed at the centre of the top circular end of a cylinder of length l = 12 cm and radius R = 9 cm as shown. The electric flux through the curved surface is nq/(5ε₀). Find n.

Question

G1 Physics · Accepted Answer

Step 1: Flux from a point charge through an open surface is qΩ/(4πε₀), where Ω is the solid angle. The curved cylinder plus bottom disk subtends the half-space below the charge; separate the bottom-disk contribution using Ωdisk=2π[1-l/√(l²+R²)]. Step 2: With l=12 cm and R=9 cm, √(l²+R²)=15 cm, so Ωdisk=2π(1-12/15)=2π/5. The total downward half-space solid angle is 2π, so Ωcurved=8π/5. Step 3: Φcurved=q(8π/5)/(4πε₀)=2q/(5ε₀), hence n=2.