Consider a uniform electric field $3\times {10}^3\widehat{\mathrm{i}} {\mathrm{NC}}^{-1}$. What is the flux through a square of side 10 cm, if the normal to its plane makes a 60⁰ angle with the X-axis?

Given, the area vector makes an angle of 60° with X-axis
Electric field, $\mathrm{E}=3\times {10}^3\widehat{i}N{C}^{-1},\theta ={60}^0$
and side of square, a =10cm
Then, area of square,
$S=(10{)}^2=100{\mathrm{cm}}^2={10}^{-2}{\mathrm{m}}^2$
Using the formula of electric flux,
$\begin{array}{l}\mathrm{\varphi }=\mathrm{E}\cdot \mathrm{S}\\ \mathrm{\varphi }=\mathrm{E}\cdot \mathrm{Scos}\mathrm{\theta }\end{array}$
$\begin{array}{l}=3\times {10}^3\times {10}^{-2}\cos {60}^{\circ }\\ =3\times 10\times \frac{1}{2}\\ =15{\mathrm{Nm}}^2{\mathrm{C}}^{-1}\end{array}$