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.

Flux, ϕ=E.→ A→=25E0i^+35E0j^. A→=E052i+3j^.0.4i^=400052×0.4=640 Nm2C−1

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The electric field in a region is given by $\bar{E} = \frac{2}{5} E_{0} \hat{i} + \frac{3}{5} E_{0} \hat{j}$ with $E_{0} = 4.0 \times 10^{3} \frac{N}{C}$ . The flux of this field through a rectangular surface area $0.4 m^{2}$ parallel to the Y-Z plane is _______ $N m^{2} C^{- 1}$ .

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answer is 640.

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Flux, $ϕ = \vec{E .} \vec{A} = [\frac{2}{5} E_{0} \hat{i} + \frac{3}{5} E_{0} \hat{j}] . \vec{A}$

$= \frac{E_{0}}{5} (2 i + 3 \hat{j}) . (0.4 \hat{i}) = \frac{4000}{5} (2 \times 0.4) = 640 N m^{2} C^{- 1}$

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