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The electric field in a region is given by E→=25E0i^+35E0j^ with E0=4.0×103NC. The flux of this field through a rectangular surface area 0.4m2 parallel to the Y-Z plane is _______ Nm2C−1

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answer is 640.

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Detailed Solution

A→=0.4i^Flux ϕ=E→ .A→ =25E0i^+35E0j^.0.4i^=0.85E0=0.85×4.0×103=640Nm2C−1

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