The electric field in a region is given by $\vec E = \frac{3}{5}E_0\hat i + \frac{4}{5}E_0\hat j$ and $E_0 = 2 \times 10^3 \, N/C$. Then,the flux of this field through a rectangular surface of area $0.2 \, m^2$ parallel to the $y-z$ plane is ...... $\frac{N \cdot m^2}{C}$.
- A$240$
- B$320$
- C$0$
- D$560$
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