An electric field is uniform, and in the positive X direction for positive X, and uniform with the same magnitude but in the negative X direction for negative X (⃗E=±100 ̂i N C⁻¹). A right circular cylinder of length 20 cm and radius 2 cm has its centre at the origin and its axis along the X-axis.…

Question

An electric field is uniform, and in the positive X direction for positive X, and uniform with the same magnitude but in the negative X direction for negative X (⃗E=±100 ̂i N C⁻¹). A right circular cylinder of length 20 cm and radius 2 cm has its centre at the origin and its axis along the X-axis. What is the flux through the side of the cylinder?

Accepted Answer

(3) 0 N m² C⁻¹. Given: ⃗E=+100 ̂i N C⁻¹ for x>0 and -100 ̂i N C⁻¹ for x<0; a cylinder of length 0.20 m and radius 0.02 m with its axis along X. Idea: φ=⃗E·Δ⃗S, so only the component of ⃗E along a surface's normal carries flux. On the curved side every area element Δ⃗S points radially outward, perpendicular to the X-axis. ⃗E is along ± X everywhere, so ⃗E⊥Δ⃗S and ⃗E·Δ⃗S=0 over the whole side. φ_side=0 N m² C⁻¹ All the flux goes through the two flat ends, each carrying Eπ r²=100×π(0.02)²≈0.126 N m² C⁻¹ outward.

Step-by-step solution

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