A uniform electric field E = 500 N/C points along +x. Compute (a) the line integral ∫E·dl from (0,0,0) to (2,0,0) metres, (b) the flux through a 2 m × 2 m square in the plane x = 1 (normal along +x), (c) the total charge in a 1 m³ box with uniform charge density ρ = 3 μC/m³. Step 1: Path is along the field: ∫E·dl = E × length = 500 × 2 = 1000 V. (This is the potential drop: V(0) − V(2) = 1000 V.) Step 2: Field is uniform and parallel to the normal: flux = E × area = 500 × 4 = 2000 N·m²/C. If the square were tilted 60° to the field, flux would be 500 × 4 × cos60° = 1000 — only the normal component counts. Step 3: Q = ρV = 3 μC/m³ × 1 m³ = 3 μC — pure accumulation, no direction. Notice the pattern: line integrals accumulate the parallel component along a path, surface integrals accumulate the normal component through a surface, volume integrals just accumulate.