Question

The electric field in a region is given by . Find the flux of this field through a rectangular surface of area 0⋅2 m² parallel to the y-z plane.

Answer 1

Given:
Electric field strength, $\overrightarrow{\mathit{E}}=\frac{3}{5}{E}_0\hat{\mathit{i}}+\frac{4}{5}{E}_0\stackrel{⏜}{j}$,
where $E_0=2.0\times {10}^3\mathrm{N}/\mathrm{C}$

The plane of the rectangular surface is parallel to the y-z plane. The normal to the plane of the rectangular surface is along the x axis. Only $\frac{3}{5}{E}_0\stackrel{⏜}{i}$ passes perpendicular to the plane; so, only this component of the field will contribute to flux.

On the other hand, $\frac{4}{5}{E}_0\stackrel{⏜}{j}$ moves parallel to the surface.
Surface area of the rectangular surface, a = 0⋅2 m²

Flux,

$$\begin{gathered} \varphi =\overrightarrow{E}.\overrightarrow{a}=E\times a \\ \varphi =\left(\frac{3}{5}\times 2\times {10}^3\right)\times \left(2\times {10}^{-1}\right){\mathrm{Nm}}^2/\mathrm{C} \\ \varphi =0.24\times {10}^3{\mathrm{Nm}}^2/\mathrm{C} \\ \varphi =240{\mathrm{Nm}}^2/\mathrm{C} \end{gathered}$$