91Ó°ÊÓ

Given figure :

Theory used :

The quantity of electric field passing through a closed surface is known as the Electric flux. Gauss's law indicates that the electric field across a surface is proportional to the angle at which it passes, hence we can determine charge inside the surface using the equation below.

${\mathrm{Φ}}_e=E·A·\cos \mathrm{θ}$ (1)

Where $\mathrm{θ}$is the angle formed between the electric field and the normal.

The electric field on surface 1 is parallel to the surface's normal but in the opposite direction, resulting in an angle of${\mathrm{θ}}_1=180°$ between the normal and the electric field.

We can get $A_1$, the area of the flat sheet by

$A_1=(4m×2m)=8m^2$

We can now insert our $E,A_1,\mathrm{and}{\mathrm{θ}}_1$values into equation (1) to get the electric flux :

$$\begin{gathered} {\mathrm{Φ}}_1=EA\cos \mathrm{θ} \\ =(400N/C)(8m^2)(\cos 180°) \\ =\boxed{\mathbf{-}\mathbf{3200}\mathbf{}\mathit{N}{\mathit{m}}^{\mathbf{2}}\mathbf{/}\mathit{C}} \end{gathered}$$

Surface 2: The electric field is perpendicular to the surface's normal, which is $90°$in this case, and $\cos 90°=0$. As a result, there is no electric flux through surface 2. That is, ${\mathrm{Φ}}_2=\boxed{\mathbf{0}}$

Surface 3: It is the same as the Surface 2. That is,${\mathrm{Φ}}_3=\boxed{\mathbf{0}}$

Surface 4: The surface's sides are $4m$long, and$\frac{2m}{\sin 30°}=4m$. In the diagram, the angle between the electric field and the normal is displayed, and it may be calculated as ${\mathrm{θ}}_4=90°-30°=60°$.

Surface 4 has an area of :

role="math" localid="1649671479054" $A_4=(4m×4m)=16m^2$

We can now put our values of$E,A_4,\mathrm{and}{\mathrm{θ}}_4$ into equation (1) to obtain the electric flux :

role="math" localid="1649671541171" $$\begin{gathered} {\mathrm{Φ}}_4=EA\cos \mathrm{θ} \\ =(400N/C)(16m^2)(\cos 60°) \\ =\boxed{\mathbf{3200}\mathbf{}\mathit{N}{\mathit{m}}^{\mathbf{2}}\mathbf{/}\mathit{C}} \end{gathered}$$

Surface 5: the electric field is perpendicular to the surface's normal, which is $90°$in this case, and $\cos 90°=0$. As a result, there is no electric flux through surface 5. That is,

${\mathrm{Φ}}_5=\boxed{\mathbf{0}}$