91Ó°ÊÓ

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

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,

$\mathrm{ϕ}=E.A=\frac{q}{{∈}_0}\left(1\right)$

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:

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

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

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

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

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

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

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

where $\stackrel{→}{E_0}=-4.00\hat{i}+6.00\hat{j}$ 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$8.23N.m^2/C$.

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

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

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