The electric field in a region is given by $\vec E = \left(\frac35E_0\hat i + \frac45E_0\hat j\right)\ \text{N C}^{-1}$. The ratio of flux of reported field through the rectangular surface of area $0.2\ \text{m}^2$ (parallel to $y$-$z$ plane) to that of the surface of area $0.3\ \text{m}^2$ (parallel to $x$-$z$ plane) is $a : b = a : 2$, where $a = $ ? [Here $\hat i$, $\hat j$ and $\hat k$ are unit vectors along $x$, $y$ and $z$-axes respectively]
Numerical value type. Enter your answer.
Answer: 1
A surface parallel to the $y$-$z$ plane has its normal along $x$: $\phi_1 = \frac35E_0\times0.2 = 0.12E_0$.
A surface parallel to the $x$-$z$ plane has its normal along $y$: $\phi_2 = \frac45E_0\times0.3 = 0.24E_0$.
$\phi_1 : \phi_2 = 1 : 2$, so $a = 1$.
Solution by Sreeraj P, M.Sc Physics