Doubtify – JEE Ki Taiyaari, Simple Bhasha Mein

❓ Question The electric field in a region is given by E = ( 2 i ^ + 4 j ^ + 6 k ^ ) × 10 3   N/C . The flux of the field through a rectangular surface parallel to the x–z plane is 6.0 N·m²·C⁻¹ . Find the area of the surface . 🖼️ Question Image ✍️ Short Solution We’ll use the electric flux formula : Φ = E ⃗ ⋅ A ⃗ = E A cos ⁡ θ Where: Φ \Phi  = Electric flux E ⃗ \vec{E}  = Electric field A ⃗ \vec{A}  = Area vector (perpendicular to the surface) θ \theta  = angle between E ⃗ \vec{E}  and A ⃗ \vec{A} 🔹 Step 1 — Determine direction of the area vector The surface is parallel to the x–z plane , ⇒ its normal vector is along the y-axis (positive or negative ĵ ). Thus, A ⃗ = A j ^​ 🔹 Step 2 — Find component of electric field perpendicular to surface Electric field: E ⃗ = ( 2 i ^ + 4 j ^ + 6 k ^ ) × 10 3 Only the ĵ-component (i.e., E y E_y E y ​ ) contributes to the flux through the x–z surface. So, E y = 4 × 10 3   N/C 🔹 Step 3...