91Ó°ÊÓ

Given figures :

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.

${\mathrm{Φ}}_e=E·A·\cos \mathrm{θ}$

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

The closed cylinder is divided into three surfaces: left, right, and cylindrical wall. The electric field is tangent to the surface of the cylindrical wall, hence the flux is zero at the cylinder wall's wall. So, ${\mathrm{Φ}}_{wall}=0$

Let's look at the flux on the left side. Because the electric field on the left side points toward the surface, the electric flux at this location is negative.

$$\begin{gathered} {\mathrm{Φ}}_{left}=E·A \\ =(-E)\left(\mathrm{π}{R}^2\right) \\ =-\mathrm{π}{R}^{2}E \end{gathered}$$

Because the electric field on the right side of the surface points outward, the electric flux at this location is positive.

$$\begin{gathered} {\mathrm{Φ}}_{right}=E·A \\ =\left(E\right)\left(\mathrm{π}{R}^2\right) \\ =\mathrm{π}{R}^{2}E \end{gathered}$$

The sum of the three fluxes through the three surfaces is the net electric flux.

$$\begin{gathered} {\mathrm{Φ}}_e={\mathrm{Φ}}_{right}+{\mathrm{Φ}}_{wall}+{\mathrm{Φ}}_{left} \\ =\mathrm{π}{R}^{2}E+0-\mathrm{π}{R}^{2}E \\ =\boxed{\mathbf{0}} \end{gathered}$$

The closed cylinder is divided into three surfaces: left, right, and cylindrical wall. The electric field is tangent to the surface of the cylindrical wall, hence the flux is zero at the cylinder wall :${\mathrm{Φ}}_{wall}=0$

Let's look at the flux on the left side. Because the electric field on the left side points outward the surface, the electric flux at this location is positive.

$$\begin{gathered} {\mathrm{Φ}}_{left}=E·A \\ =\left(E\right)\left(\mathrm{π}{R}^2\right) \\ =\mathrm{π}{R}^{2}E \end{gathered}$$

Because the electric field on the right side of the surface points outward, the electric flux at this location is positive.

$$\begin{gathered} {\mathrm{Φ}}_{left}=E·A \\ =\left(E\right)\left(\mathrm{π}{R}^2\right) \\ =\mathrm{π}{R}^{2}E \end{gathered}$$

The sum of the three fluxes through the three surfaces is the net electric flux.

$$\begin{gathered} {\mathrm{Φ}}_e={\mathrm{Φ}}_{right}+{\mathrm{Φ}}_{wall}+{\mathrm{Φ}}_{left} \\ =\mathrm{π}{R}^{2}E+0+\mathrm{π}{R}^{2}E \\ =\boxed{\mathbf{2}\mathit{π}{\mathit{R}}^{\mathbf{2}}\mathit{E}} \end{gathered}$$