Question

An electric field vector E = 4xi - (y² + 1)jN/C passes through the box shown in figure. The flux of the electric field through surfaces ABCD and BCGF are marked as ϕ_I and ϕ_II respectively. The difference between (ϕ_I – ϕ_II) is (in Nm²/C) ___________. 

Answer 1

The flux passes through ABCD (x – y) plane is zero, because electric field parallel to surface. Flux of the electric field through surface BCGF (y - z)

At BCGF (electric field) 

vector E = 12i - (y² + 1)j

(x = 3m)

Flux ϕ_II = 12 × 4 = 48 Nm²/C

So ϕ_I - ϕ_II = 0 – 48 = – 48 Nm²/C

Answer 2

Flux via ABCD

Flux via BCEF