The electric field in a region is given by E- - 3-5 Eo i - 4-5 Eo j with Eo - 2-0 - 103 N-C- Find the flux of this field in Nm2C-1 through a rectangular surface of area 0-2 m 2 parallel to the Y-Z plane-

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The electric field in a region is given by $\to E = \frac{3}{5} E_{o} ˆ i + \frac{4}{5} E_{o} ˆ j$ with $E_{o} = 2.0 \times 10^{3}$ N/C. Find the flux of this field (in $N m^{2} C^{- 1}$ ) through a rectangular surface of area 0.2 m $^{2}$ parallel to the Y-Z plane.

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\to E = \frac{3}{5} E_{0} \to i + \frac{4}{5} E_{0}^j

E_{0} = 2.0 \times 10^{3} N C^{- 1}

0.2 m^{2}

y - z

\to E = \frac{2}{5} E_{0}^i + \frac{3}{5} E_{0}^j

E_{0} = 4.0 \times 10^{3} N/C

0.4 m^{2}

({in Nm}^{2} C^{- 1})

\to E = (1200^i + 1600^j) N/C

0.2 m^{2}

y - z

\vec{E} = \frac{3}{5} E_{0} \vec{i} + \frac{4}{5} E_{0} \vec{j} with E_{0} = 2 \cdot 0 \times 10^{3} N C^{- 1}

\to E = \frac{3}{5} E_{0}^j

E_{0} = 2 \times 10^{3} N C^{- 1}

0.2 m^{2}

Y - Z

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