Question

The electric field in a region is given by 

${\displaystyle \overrightarrow{\text{E}}=\frac{3}{5}{\text{E}}_0\overrightarrow{\text{i}}+\frac{4}{5}{\text{E}}_0\overrightarrow{\text{i}} \text{with}{\text{E}}_0=2.0\times {10}^3\text{N}{\text{C}}^{-1}.}$ 

 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 ${\displaystyle \overrightarrow{\text{E}}=\frac{3}{5}{\text{E}}_0\hat{\text{i}}+\frac{4}{5}{\text{E}}_0}$ $$\stackrel{⌢}{j}$$,

where E₀ = 2.0  10³ N/C

he 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 ${\displaystyle \frac{3}{5}{\text{E}}_0}$$$\stackrel{⌢}{i}$$, passes perpendicular to the plane; so, only this component of the field will contribute to flux.
On the other hand, ${\displaystyle \frac{4}{5}{\text{E}}_0}$$$\stackrel{⌢}{j}$$  moves parallel to the surface.
Surface area of the rectangular surface, a = 0⋅2 m²

Flux,

${\displaystyle \varphi =\overrightarrow{\text{E}}.\overrightarrow{\text{a}}=\text{E}\times \text{a}}$

${\displaystyle \varphi =(\frac{3}{5}\times 2\times {10}^3)\times (2\times {10}^{-1})\text{N}{\text{m}}^2/\text{C}}$

${\displaystyle \varphi =0.24\times {10}^3\text{N}{\text{m}}^2/\text{C}}$

${\displaystyle \varphi =240\text{N}{\text{m}}^2/\text{C}}$