The electric field in a region is given by E→=35E0i^+45E0j^ with E0=2×103 N/C. Find the flux of this field through a rectangular surface of area 0.2 m2 parallel to Y–Z plane.

ϕ=E¯⋅A¯=35E0×0.2=0.65×2×103=240 Nm2/C

The electric field in a region is given by E→=35E0i^+45E0j^ with E0=2×103 N/C. Find the flux of this field through a rectangular surface of area 0.2 m2 parallel to Y–Z plane.

ϕ=E¯⋅A¯=35E0×0.2=0.65×2×103=240 Nm2/C

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The electric field in a region is given by $\vec{E} = \frac{3}{5} E_{0} \hat{i} + \frac{4}{5} E_{0} \hat{j}$ with $E_{0} = 2 \times 10^{3} N / C$ . Find the flux of this field through a rectangular surface of area 0.2 m² parallel to Y–Z plane.

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$240 {Nm}^{2} / C$

$40 {Nm}^{2} / C$

$340 {Nm}^{2} / C$

$140 {Nm}^{2} / C$

answer is A.

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$\begin{aligned} ϕ = \bar{E} \cdot \bar{A} = \frac{3}{5} E_{0} \times 0 .2 = \frac{0 .6}{5} \times 2 \times 10^{3} \\ = 240 {Nm}^{2} / C \end{aligned}$

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The electric field in a region is given by E→=35E0i^+45E0j^ with E0=2×103 N/C. Find the flux of this field through a rectangular surface of area 0.2 m2 parallel to Y–Z plane.

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The electric field in a region is given by $\vec{E} = \frac{3}{5} E_{0} \hat{i} + \frac{4}{5} E_{0} \hat{j}$ with $E_{0} = 2 \times 10^{3} N / C$ . Find the flux of this field through a rectangular surface of area 0.2 m² parallel to Y–Z plane.