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 $E_0 = 2\times10^3\, N/C$. Then, the flux of this field through a rectangular surface of area $0.2\, m^2$ parallel to the $y-z$ plane is......$\frac{{N - {m^2}}}{C}$

  • A

    $240$

  • B

    $320$

  • C

    $0$

  • D

    $560$

Similar Questions

$\mathrm{C}_1$ and $\mathrm{C}_2$ are two hollow concentric cubes enclosing charges $2 Q$ and $3 Q$ respectively as shown in figure. The ratio of electric flux passing through $\mathrm{C}_1$ and $\mathrm{C}_2$ is :

  • [JEE MAIN 2024]

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