Question

Fig. 23-31 shows a Gaussian surface in the shape of a cube with edge length 1.40m. What are (a) the net flux through the surface and (b) the net charge$q_{enc}$enclosed by the surface if $\overrightarrow{E}=3.00y\hat{j}+\overrightarrow{E_0}$with yin meters? What are (c)$\varphi$and (d) $q_{enc}$if $\overrightarrow{E}=\left[-4.00\hat{i}+\left(6.00+3.00y\right)\hat{j}\right]\raisebox{1ex}{$N$}\!\left/ \!\raisebox{-1ex}{$C$}\right.$?

Short answer

1. The net flux through the surface is $8.23N.\raisebox{1ex}{$m^2$}\!\left/ \!\raisebox{-1ex}{$C$}\right.$ .
2. The net enclosed charge by the surface is $7.29\times {10}^{-11}C$.
3. The net flux through the surface is $8.23N\raisebox{1ex}{$m^2$}\!\left/ \!\raisebox{-1ex}{$C$}\right.$ .
4. The net enclosed charge is $7.29\times {10}^{-11}C$.

Step-by-step solution

The given data

The edge length of the cube,$a=1.4m$

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,

$\varphi =E.A=\frac{q}{{\in }_0}\left(1\right)$

a) Calculation of the net flux 

Let the area of the cube,$A={\left(1.40m\right)}^2$

Then, the net flux enclosed by the cube is given using equation (1) as:

$$\begin{gathered} \varphi =\left(3.00y\hat{j}\right).\left(-A\hat{j}\right){\left|{}_{y=0m}+\left(3.00y\hat{j}\right).\left(A\hat{j}\right)\right|}_{y=1.40m} \\ \varphi =\left(3.00\raisebox{1ex}{$N$}\!\left/ \!\raisebox{-1ex}{$C$}\right.\right){\left(1.40m\right)}^2 \\ \varphi =8.23N.m^2/C \end{gathered}$$

Hence, the value of the flux is$8.23N.m^2/C$ .

b) Calculation of the net charge enclosed by the cube 

The charge enclosed by the cube is now given using equation (i) as:

$$\begin{gathered} q_{enc}=\left(8.85\times {10}^{-12}{C}^2/N{m}^2\right)\left(8.23N.m^2/C\right) \\ {q}_{enc}=7.29\times {10}^{-11}C \end{gathered}$$

Hence, the value of the charge is $7.29\times {10}^{-11}C$.

c) Calculation of the net flux 

Let the area of the cube,$A={\left(1.40m\right)}^2$

The electric field can be re-written as:$\overrightarrow{E}=3.00y\hat{j}+\overrightarrow{E_0}$,

where $\overrightarrow{E_0}=-4.00\hat{i}+6.00\hat{j}$ is a constant field which does not contribute to the net flux through the cube.

Thus, the net flux remains constant for this field and is$8.23N.m^2/C$.

d) Calculation of the net charge enclosed by the cube

The charge enclosed by the cube is now given using equation (1) such that,

$$\begin{gathered} q_{enc}=\left(8.85\times {10}^{-12}{C}^2/N{m}^2\right)\left(8.23N.m^2/C\right) \\ {q}_{enc}=7.29\times {10}^{-11}C \end{gathered}$$

Hence, the value of the charge is$7.29\times {10}^{-11}C$ .