What net charge is enclosed by the Gaussian cube of Problem 2?

(a) The electric field is given by:$\overrightarrow{\mathrm{E}}=4.0\hat{\mathrm{i}}-3.0({\mathrm{y}}^2+2.0)\hat{\mathrm{j}}$

(b) Gaussian cube with edge length,$\mathrm{a}=2.0\hspace{0.33em}\mathrm{m}$

Using the concept of the electric flux from the Gauss flux theorem, we can get the net charge within the surface due to the net flux leaving from all the surfaces.

Formula:

The electric flux passing through the surface enclosed within the volume,

$\mathrm{\varphi }=\int \overrightarrow{\mathrm{E}}\cdot \left(\overrightarrow{\mathrm{dA}}\right)=\mathrm{q}/{\mathrm{\epsilon }}_0$ (i)

The side length of the cube is given as:$a=2.0\hspace{0.33em}m$.

On the top face of the cube$y=2.0\hspace{0.33em}m$ and$\overrightarrow{dA}=\left(dA\right)\hat{j}$.

Thus, the value of the electric field using this value is given as:

role="math" localid="1657517368708" $$\begin{gathered} \overrightarrow{\mathrm{E}}=4.0\hat{\mathrm{i}}-3.0(2^2+2.0)\hat{\mathrm{j}} \\ =4\hat{\mathrm{i}}-18\hat{\mathrm{j}} \end{gathered}$$

Now, using equation (i), we can get the flux through this surface is given as:

role="math" localid="1657517458193" $$\begin{gathered} \mathrm{\varphi }=\underset{\mathrm{top}}{\int }\left(4\hat{\mathrm{i}}-18\hat{\mathrm{j}}\right).\left(\mathrm{dA}\right)\hat{\mathrm{j}} \\ =-18\underset{\mathrm{top}}{\int }\mathrm{da} \\ =(-18){(2.0)}^2\mathrm{N}.{\mathrm{m}}^2/\mathrm{C} \\ =-72\mathrm{N}.{\mathrm{m}}^2/\mathrm{C} \end{gathered}$$

On the bottom face of the cube$y=0$ and$\overrightarrow{dA}=\left(dA\right)(-\hat{j})$.

Thus, the value of the electric field using this value is given as:

$$\begin{gathered} \overrightarrow{\mathrm{E}}=4.0\hat{\mathrm{i}}-3.0(0^2+2.0)\hat{\mathrm{j}} \\ =4\hat{\mathrm{i}}-6\hat{\mathrm{j}} \end{gathered}$$

Now, using equation (i), we can get the flux through this surface is given as:

role="math" localid="1657517648740" $$\begin{gathered} \mathrm{\varphi }=\underset{bot}{\int }\left(4\hat{\mathrm{i}}-18\hat{\mathrm{j}}\right).\left(\mathrm{dA}\right)\hat{\mathrm{j}} \\ =6\int \mathrm{da} \\ =6{(2.0)}^2\mathrm{N}.{\mathrm{m}}^2/\mathrm{C} \\ =+24\mathrm{N}.{\mathrm{m}}^2/\mathrm{C} \end{gathered}$$

On the left face of the cube,$\overrightarrow{dA}=\left(dA\right)(-\hat{i})$.

Now, using equation (i), we can get the flux through this surface is given as:

$$\begin{gathered} \mathrm{\varphi }=\underset{left}{\int }\left(4\hat{\mathrm{i}}+E_y\hat{\mathrm{j}}\right).\left(\mathrm{dA}\right)\left(-\hat{i}\right) \\ =-4\underset{left}{\int }\mathrm{da} \\ =-4{(2.0)}^2\mathrm{N}.{\mathrm{m}}^2/\mathrm{C} \\ =-16\mathrm{N}.{\mathrm{m}}^2/\mathrm{C} \end{gathered}$$

On the back face of the cuberole="math" localid="1657516163117" $\overrightarrow{dA}=\left(dA\right)(-\hat{k})$.

But since E has no z component,$\overrightarrow{E}\cdot \overrightarrow{dA}=0$.

Now, using equation (i), we can get the flux through this surface is given as:$\varphi =0$

The flux through the front face is zero, while that through the right face is the opposite of that through the left one, or.$+16\hspace{0.33em}\mathrm{N}·{\mathrm{m}}^2/\mathrm{C}$Thus the net flux through the cube is given as:

$$\begin{gathered} \mathrm{\varphi }=\left(-72+24-16+0+0+16\right)\mathrm{N}·{\mathrm{m}}^2/\mathrm{C} \\ =-48\mathrm{N}·{\mathrm{m}}^2/\mathrm{C} \end{gathered}$$

Thus, the net enclosed charge q is given using equation (i) as follows:

$$\begin{gathered} \mathrm{q}=\left(8.85\times {10}^{-12}{\mathrm{C}}^2/\mathrm{N}.{\mathrm{m}}^2\right)\left(-48\mathrm{N}.{\mathrm{m}}^2/\mathrm{C}\right) \\ =-4.2\times {10}^{-10}\hspace{0.33em}\mathrm{C} \end{gathered}$$

Hence, the value of the charge is role="math" localid="1657515548344" $-4.2\times {10}^{-10}\hspace{0.33em}\mathrm{C}$.