The electric field in a region is E ⃗ \vec{E} E = (2E₀/5) i ^ \hat{i} i ^ + (3E₀/5) j ^ \hat{j} j ^ , where E₀ = 4.0 × 10³ N C⁻¹. Find the flux through a rectangular surface of area 0.4 m² parallel to the Y-Z plane, in N m² C⁻¹.
Physics · Electrostatic
JEE Mains 2021

Question

The electric field in a region is E ⃗ \vec{E} E = (2E₀/5) i ^ \hat{i} i ^ + (3E₀/5) j ^ \hat{j} j ^ , where E₀ = 4.0 × 10³ N C⁻¹. Find the flux through a rectangular surface of area 0.4 m² parallel to the Y-Z plane, in N m² C⁻¹.

Physics · Electrostatic

JEE Mains 2021

G1 Physics · Accepted Answer

Step 1: The surface is parallel to the yz-plane, so its area vector is along x and only Eₓ contributes. Step 2: Eₓ=2E₀/5=(2/5)(4.0×10³)=1600 N C⁻¹. Step 3: Φ=EₓA=1600×0.4=640 N m² C⁻¹.

The electric field in a region is E ⃗ \vec{E} E = (2E₀/5) i ^ \hat{i} i ^ + (3E₀/5) j ^ \hat{j} j ^ ​ , where E₀ = 4.0 × 10³ N C⁻¹. Find the flux through a rectangular surface of area 0.4 m² parallel to the Y-Z plane, in N m² C⁻¹.

The electric field in a region is E ⃗ \vec{E} E = (2E₀/5) i ^ \hat{i} i ^ + (3E₀/5) j ^ \hat{j} j ^ , where E₀ = 4.0 × 10³ N C⁻¹. Find the flux through a rectangular surface of area 0.4 m² parallel to the Y-Z plane, in N m² C⁻¹.