Figure 23-35 shows a closed Gaussian surface in the shape of a cube of edge length$\mathbf{2}\mathbf{.00}\mathbf{}\mathbf{\text{m}}\mathbf{,}$with ONE corner at${\mathit{x}}_{\mathbf{1}}\mathbf{=}\mathbf{5.00}\mathbf{}\mathbf{\text{m}}$,${\mathbf{y}}_{\mathbf{1}}\mathbf{=}\mathbf{4}\mathbf{.00}\mathbf{\text{m}}$.The cube lies in a region where the electric field vector is given by$\overrightarrow{\mathbf{E}}\mathbf{=}\mathbf{[}\mathbf{-}\mathbf{3}\mathbf{.00}\mathbf{}\widehat{\mathbf{i}}\mathbf{-}\mathbf{4}\mathbf{.00}{\mathbf{y}}^{\mathbf{2}}\widehat{\mathbf{j}}\mathbf{+}\mathbf{3}\mathbf{.00}\widehat{\mathbf{k}}\mathbf{]}\mathbf{}\mathbf{\text{N/C}}$with yin meters. What is the net charge contained by the cube?

The electric field, $\overrightarrow{\mathrm{E}}=[-3.00\widehat{\mathrm{i}}-4.00{\mathrm{y}}^2\widehat{\mathrm{j}}+3.00\widehat{\mathrm{k}}]\text{\hspace{0.17em}N/C}$

The edge length of the cube is$\mathrm{a}=2.00\text{\hspace{0.17em}m}$ with one corner at${\mathrm{x}}_1=5.00\text{\hspace{0.17em}m}$ ,${\mathrm{y}}_1=4.00\text{\hspace{0.17em}m}$

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,

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

None of the constant terms will result in a nonzero contribution to the flux, so we focus on the x-dependent term only:

${\mathrm{E}}_{\mathrm{non}-\mathrm{constant}}=-4.00{\mathrm{y}}^2\widehat{\mathrm{i}}$

The face of the cube located at $\mathrm{y}=4.00\mathrm{m}$has an area $\mathrm{A}=4.00{\mathrm{m}}^2$(and it “faces” the +j direction) and has a “contribution” to the flux that is given using equation (1) as:

$\begin{array}{c}{\mathrm{E}}_{\mathrm{non}-\mathrm{constant}}\mathrm{A}=(3.00\mathrm{N}/\mathrm{C})(-4.00\times {\left(4.00\mathrm{m}\right)}^2)\\ =-256\text{\hspace{0.17em}N}\cdot {\text{m}}^{\text{2}}\text{/C}\end{array}$

The face of the cube located at $\mathrm{y}=2.00\mathrm{m}$has the same area A (however, this one “faces” the –j direction) and a contribution to the flux that is given using equation (1) as:

$\begin{array}{c}-{\mathrm{E}}_{\mathrm{non}-\mathrm{constant}}\mathrm{A}=(-4.00\mathrm{N}/\mathrm{C})(-4.00\times {\left(2.00\mathrm{m}\right)}^2)\\ =64\text{\hspace{0.17em}N}\cdot {\text{m}}^{\text{2}}\text{/C}\end{array}$

Thus, the net flux is given using equations (a) and (b) as given:

$\begin{array}{c}\mathrm{\varphi }=(-256+64)\text{\hspace{0.17em}N}\cdot {\text{m}}^{\text{2}}\text{/C}\\ =-192\text{\hspace{0.17em}N}\cdot {\text{m}}^{\text{2}}\text{/C}\end{array}$.

According to Gauss’s law, the net charge contained by the face is given as:

$\begin{array}{c}{\mathrm{q}}_{\mathrm{enc}}=(8.85\times {10}^{-12}{\text{C}}^{\text{2}}/\text{N}\cdot {\text{m}}^{\text{2}})(-192\text{N}\cdot {\text{m}}^{\text{2}}\text{/C})\\ =-1.70\times {10}^{-9}\text{C}\end{array}$

Hence, the value of the charge is $-1.70\times {10}^{-9}\text{C}$.