In a certain region, the electric field is expressed as E = 3 5 E₀ i + 4 5 E₀ j , where E₀ = 2.0 10^3 N C ⁻¹ . Determine the flux of this field through a rectangular surface having an area of 0.2 m ^2 that is parallel to the y – z plane.

Step-by-step solution

The surface is parallel to the y - z plane, so its area vector points along the x -axis. A = 0.2 i m ^2 The electric field is given as: E = 3 5 E₀ i + 4 5 E₀ j The electric flux through the surface is the dot product of the electric field and the area vector: = E A = ( 3 5 E₀ i + 4 5 E₀ j ) (0.2 i ) = 3 5 E₀ 0.2 Substituting E₀ = 2.0 10^3 N C ⁻¹ : = 3 5 2000 0.2 = 1200 0.2 = 240 N m ^2 C ⁻¹