A $1.0cm\times 1.0cm\times 1.0cm$ box with its edges aligned with the $xyz$-axes is in the electric field $\overrightarrow{E}=(350x+150)\hat{i}N/C$, where x is in meters. What is the net electric flux through the box?

Given :

Dimensions of the box : $1.0cm\times 1.0cm\times 1.0cm$

The electric field is : $\overrightarrow{E}=(350x+150)\hat{i}N/C$

Theory used :

The net electric flux through the surface of a box is directly proportional to the magnitude of the net charge enclosed by the box. The net electric flux due to a point charge inside a box is independent of box's size, only depends on net amount of charge enclosed.

The field points along the $x$-axis so it is parallel to all the four faces that encircle it and it's perpendicular to the two faces as in the figure bellow.

Therefore, all of the four faces that are parallel to the field don't contribute to net flux, only the two faces perpendicular to it do. One of them is $x=0$so :

$\overrightarrow{E}=150\overrightarrow{i}N/C$

The other one is $x=1.0cm=0.01m$, so there are two possibilities.

We get :

$$\begin{gathered} \overrightarrow{E}=(350·0.01+150)\overrightarrow{i}N/C \\ =153\overrightarrow{i}N/C \end{gathered}$$

by plugging in the specified numerical numbers.

Now, ${\overrightarrow{E}}_1$points inside, so the face at$x=0$contributes negatively, while ${\overrightarrow{E}}_2$points outside, so the face at $x=1.0cm$contributes positively.

So, we have

role="math" localid="1649665790231" $$\begin{gathered} \Phi =-{\overrightarrow{E}}_1(0.01m)²+{\overrightarrow{E}}_2(0.01m)² \\ =(0.01m)²(E₂-E₁). \\ \Rightarrow \Phi =\boxed{\mathbf{3}\mathbf{.}\mathbf{5}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{4}\mathbf{}}\mathit{N}\mathit{m}\mathbf{/}\mathit{C}} \end{gathered}$$