Question

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?

Short answer

The net charge contained by the cube is $-1.70\times {10}^{-9}\text{C}$.

Step-by-step solution

The given data 

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${\mathrm{x}}_1=5.00\text{\hspace{0.17em}m}$ ,${\mathrm{y}}_1=4.00\text{\hspace{0.17em}m}$

Understanding the concept of Gauss law-planar symmetry 

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)

Calculation of the net charge 

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 $-1.70\times {10}^{-9}\text{C}$.