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Question

The electric field in a region is given by E=35E0 i+45 E0 j with E0=2·0×103 N C-1. Find the flux of this field through a rectangular surface of area 0⋅2 m2 parallel to the y-z plane.

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Solution

Given:
Electric field strength, E=35E0i^+45 E0j,
where E0 = 2.0×103 N/C

The plane of the rectangular surface is parallel to the y-z plane. The normal to the plane of the rectangular surface is along the x axis. Only 35E0i passes perpendicular to the plane; so, only this component of the field will contribute to flux.

On the other hand, 45E0j moves parallel to the surface.
Surface area of the rectangular surface, a = 0⋅2 m2

Flux,

ϕ=E.a =E×aϕ= 35×2×103×2×10-1 Nm2/Cϕ=0.24×103 Nm2/Cϕ=240 Nm2/C

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