Figure 23-34 shows a closed Gaussian surface in the shape of a cube of edge length 2.00 m. It lies in a region where the non-uniform electric field is given by $\overrightarrow{\mathbf{E}}\mathbf{=}\left[(3.00x+4.00)\hat{i}+6.00\hat{j}+7.00\hat{k}\right]\mathbf{N}\mathbf{/}\mathbf{C}$, with xin meters. What is the net charge contained by the cube?

Using the gauss flux theorem, we can get the net flux through the surfaces. Now, using the same concept, we can get the net charge contained in the cube.

Formula:

The electric flux passing through the surface,

$\mathrm{\varphi }=\mathrm{E}.\mathrm{A}=\frac{\mathrm{q}}{{\mathrm{\epsilon }}_0}$ (1)

None of the constant terms will result in a nonzero contribution to the flux, so we focus on the x-dependent term only. In Si units, we have

${\mathrm{E}}_{\mathrm{non}-\mathrm{constant}}=3\mathrm{x}\hat{\mathrm{i}}$

The face of the cube located at x = 0 (in the y-z plane) has area $\mathrm{A}=4{\mathrm{m}}^2$(and it “faces” the +i direction) and has a “contribution” to the flux that is given using equation (1) such that,

$$\begin{gathered} {{\mathrm{E}}_{\mathrm{non}-\mathrm{constant}}}^{\mathrm{A}}=\left(\right(3.00\left)\right(0)\mathrm{N}/\mathrm{C}(4{\mathrm{m}}^2) \\ =0 \end{gathered}$$

The face of the cube located at x = -2m has the same area A (and this one “faces” the –i direction) and a contribution to the flux that is given using equation (i) as:

$$\begin{gathered} {{\mathrm{E}}_{\mathrm{non}-\mathrm{constant}}}^{\mathrm{A}}=-\left(\right(3.00\left)\right(-2)\mathrm{N}/\mathrm{C}(4{\mathrm{m}}^2) \\ =24\mathrm{N}.{\mathrm{m}}^2/\mathrm{C} \end{gathered}$$

Thus, the net flux is given by:

$$\begin{gathered} \mathrm{\varphi }=0\mathrm{N}.{\mathrm{m}}^2/\mathrm{C}+24\mathrm{N}.{\mathrm{m}}^2/\mathrm{C} \\ =24\mathrm{N}.{\mathrm{m}}^2/\mathrm{C} \end{gathered}$$

According to Gauss’ law, the net enclosed charge by the cube is given using equation (1) as:

$$\begin{gathered} {\mathrm{q}}_{\mathrm{enc}}=(8.85\times {10}^{-12}\mathrm{N}.{\mathrm{m}}^2/{\mathrm{C}}^2)(24\mathrm{N}.{\mathrm{m}}^2/\mathrm{C}) \\ =2.13\times {10}^{-10}\mathrm{C} \end{gathered}$$

Hence, the value of the charge is $2.13\times {10}^{-10}\mathrm{C}$.