91Ó°ÊÓ

1. Electric field,$\mathrm{E}=4.0\hat{\mathrm{i}}-3.0\left({\mathrm{y}}^2+2.0\right)\hat{\mathrm{j}}$
2. Edge length of the Gaussian cube, a=2.0 m

Using the concept of the Gauss flux theorem, we can get the flux passing through the surface as per the given electric field. Now, the summation of all values will give the net flux through the volume.

Formula:

The electric flux through an enclosed surface, $\mathrm{ϕ}=∫\mathrm{E}.\mathrm{dA}$ (1)

The side length of the cube is given as: $\mathrm{a}=2.0\mathrm{m}$

On the top face of the cube $\mathrm{y}=2.0\mathrm{m}$$\mathrm{and}\overline{\mathrm{d}}\stackrel{→}{\mathrm{A}}=\left(\mathrm{dA}\right)\widehat{\hat{\mathrm{j}}}$

Thus, we have the electric field by this value as:

$\begin{array}{l}\mathrm{E}=4.0\hat{\mathrm{i}}-3.0\left(2^2+2.0\right)=\hat{\mathrm{j}}\\ =4\hat{\mathrm{i}}-18\hat{\mathrm{j}}\end{array}$

Now, the flux is given using equation (1) as:

$$\begin{gathered} \mathrm{ϕ}={∫}_{\mathrm{top}}\left(4\hat{\mathrm{i}}-18\hat{\mathrm{j}}\right).\left(\mathrm{dA}\right)\hat{\mathrm{j}} \\ =-18{∫}_{\mathrm{top}}\mathrm{dA} \\ =\left(-18\right){\left(2.0\right)}^2\mathrm{N}.{\mathrm{m}}^2/\mathrm{C} \\ =-72\mathrm{N}.{\mathrm{m}}^2/\mathrm{C} \end{gathered}$$

Hence, the value of the flux is $-72\mathrm{N}.{\mathrm{m}}^2/\mathrm{C}$.

On the bottom face of the cube y=0 m and$\overline{\mathrm{d}}\stackrel{→}{\mathrm{A}}=\left(\mathrm{dA}\right)\left(-\hat{\mathrm{j}}\right)$.

Now, we have the electric field using this value as:

$\begin{array}{l}\mathrm{E}=4.0\hat{\mathrm{i}}-3.0\left(0^2+20\right)\hat{\mathrm{j}}\\ =4\hat{\mathrm{i}}-6\hat{\mathrm{j}}\end{array}$

Thus, the flux is given using equation (1) as:

$$\begin{gathered} \mathrm{ϕ}={∫}_{\mathrm{bottom}}\left(4\hat{\mathrm{i}}-6\hat{\mathrm{j}}\right).\left(\mathrm{dA}\right)\left(-\hat{\mathrm{j}}\right) \\ =6{∫}_{\mathrm{bottom}}\mathrm{dA} \\ =6{(20.)}^2\mathrm{N}.\mathrm{m}/\mathrm{C} \\ =+24\mathrm{N}.{\mathrm{m}}^2/\mathrm{C} \end{gathered}$$

Hence, the value of the flux is $+24\mathrm{N}.{\mathrm{m}}^2/\mathrm{C}$.

On the left face of the cube,$\stackrel{→}{\mathrm{dA}}=\left(\mathrm{dA}\right)\left(-\hat{\mathrm{i}}\right)$.

Thus, the value of the electric flux is given using equation (1) as:

$$\begin{gathered} \mathrm{ϕ}={∫}_{\mathrm{left}}\left(4\hat{\mathrm{i}}-{\mathrm{E}}_{\mathrm{y}}\hat{\mathrm{j}}\right).\left(\mathrm{dA}\right)\left(-\hat{\mathrm{i}}\right) \\ =-4{∫}_{\mathrm{left}}\mathrm{dA} \\ =4{(20.)}^2\mathrm{N}.\mathrm{m}/\mathrm{C} \\ =-16\mathrm{N}.{\mathrm{m}}^2/\mathrm{C} \end{gathered}$$

Hence, the value of the electric flux is $-16\mathrm{N}.{\mathrm{m}}^2/\mathrm{C}$.

On the back face of the cube, $\stackrel{→}{\mathrm{dA}}=\left(\mathrm{dA}\right)\left(-\hat{\mathrm{k}}\right)$.

But since E has no z-component, thus, using equation (1), the flux can be given as:

$$\begin{gathered} \mathrm{Ï•}=\mathrm{E}.\mathrm{dA} \\ =0 \end{gathered}$$

Thus, the value of the electric field is zero.

We now have to add the flux through all six faces. One can easily verify that

the flux through the front face is zero, while that through the right face is the opposite of that through the left one, or $+16\mathrm{N}.{\mathrm{m}}^2/\mathrm{C}$.

Thus the net flux through the cube is given as:

$$\begin{gathered} \mathrm{Ï•}=\left(-72+24-16+0+0+16\right)\mathrm{N}.{\mathrm{m}}^2/\mathrm{C} \\ =-48\mathrm{N}.{\mathrm{m}}^2/\mathrm{C} \end{gathered}$$

Hence, the value of the flux is $-48\mathrm{N}.{\mathrm{m}}^2/\mathrm{C}$.