Question

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?

Short answer

The net charge contained by the cube is $2.13\times {10}^{-10}\mathrm{C}$.

Step-by-step solution

The given data

1. The given electric field,$\overrightarrow{\mathrm{E}}=\left[(3.00\mathrm{x}+4.00)\hat{\mathrm{i}}+6.00\hat{\mathrm{j}}+7.00\hat{\mathrm{k}}\right]\mathrm{N}/\mathrm{C}$

1. Edge length of the cube, a = 2.00 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,

$\mathrm{\varphi }=\mathrm{E}.\mathrm{A}=\frac{\mathrm{q}}{{\mathrm{\epsilon }}_0}$ (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. 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 $2.13\times {10}^{-10}\mathrm{C}$.