The electric field in a region is given by $$\vec{E} = \left(\frac{3}{5}E_0\hat{i} + \frac{4}{5}E_0\hat{j}\right)$$ N C$$^{-1}$$. The ratio of flux of reported field through the rectangular surface of area 0.2 m$$^2$$ (parallel to $$y-z$$ plane) to that of the surface of area 0.3 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]

The electric field is given as $$\vec{E} = \frac{3}{5}E_0\hat{i} + \frac{4}{5}E_0\hat{j}$$.

For the first surface of area $$A_1 = 0.2\,\text{m}^2$$ parallel to the $$y\text{-}z$$ plane, the outward normal is along the $$x$$-direction ($$\hat{i}$$). The flux through this surface is $$\phi_1 = \vec{E} \cdot \hat{i} \times A_1 = \frac{3}{5}E_0 \times 0.2 = \frac{3E_0}{25}$$.

For the second surface of area $$A_2 = 0.3\,\text{m}^2$$ parallel to the $$x\text{-}z$$ plane, the outward normal is along the $$y$$-direction ($$\hat{j}$$). The flux through this surface is $$\phi_2 = \vec{E} \cdot \hat{j} \times A_2 = \frac{4}{5}E_0 \times 0.3 = \frac{12E_0}{50} = \frac{6E_0}{25}$$.

The ratio of the two fluxes is $$\frac{\phi_1}{\phi_2} = \frac{3E_0/25}{6E_0/25} = \frac{3}{6} = \frac{1}{2}$$.

This gives the ratio $$a : b = 1 : 2$$, and since $$b = 2$$, we get $$a = 1$$.