The electric field in a region is given by $\overrightarrow{\mathrm{E}}=4\widehat{\mathrm{i} }+10\hat{\mathrm{j}} \mathrm{N}/\mathrm{C}$. The flux of this field through a square of side 10 cm whose plane is parallel to the XZ plane.

Given: $\overrightarrow{\mathrm{E}}=4\widehat{\mathrm{i}}+10\hat{\mathrm{j}} \mathrm{N}/\mathrm{C}; l=10 \mathrm{cm}=\frac{10}{100}\mathrm{m}=0.1 \mathrm{m}$

$\therefore$ area: $A=l^2$  $\Rightarrow A={(0.1)}^2=1\times {10}^{-2}{m}^2$

Flux through the square whose plane is parallel to the $xz$ plane,

Flux, $\mathrm{\varphi }=\overrightarrow{\mathrm{E}}.\overrightarrow{\mathrm{A}}=(4\hat{\mathrm{i}}+10\hat{\mathrm{j}}).(1\times {10}^{-2})\hat{\mathrm{j}}=10\times {10}^{-2}=0.1 \mathrm{Vm}$