What is the net electric flux through the two cylinders shown inFIGURE EX24.16? Give your answer in terms of $R\mathrm{and}E$

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.

${\Phi }_e=E·A·\cos \theta$

Where $\theta$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, ${\Phi }_{wall}=0$

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.

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

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

$$\begin{gathered} {\Phi }_{right}=E·A \\ =\left(E\right)\left(\pi R^2\right) \\ =\pi R^{2}E \end{gathered}$$

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

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

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 :${\Phi }_{wall}=0$

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.

$$\begin{gathered} {\Phi }_{left}=E·A \\ =\left(E\right)\left(\pi R^2\right) \\ =\pi R^{2}E \end{gathered}$$

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

$$\begin{gathered} {\Phi }_{left}=E·A \\ =\left(E\right)\left(\pi R^2\right) \\ =\pi R^{2}E \end{gathered}$$

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

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