Consider an electric field E→=E0x^, where E0 is a constant. The flux through the area (as shown in the figure) due to this field is :
2E0a2
E0a2
2E0a22
Here, E→=E0x^In this problem, x^, y^ and z^ are the unit vectors along x, y and z axes respectivelyShaded area, A→=PQ→×QR→A→=(0x^+ay^+0z^)×(ax^+0y^+az^)ϕ=E→⋅A→=E0x^⋅a2x^−a2z^=E0a2