91影视

1. Net charge enclosed by the surface,${\mathrm{q}}_{\mathrm{enc}}=+24{\mathrm{蔚}}_0\mathrm{C}$
2. Electric field, $\stackrel{鈫�}{\mathrm{E}}=[(10.0+2.00)\hat{\mathrm{j}}+\mathrm{bz}\hat{\mathrm{k}}]\mathrm{N}/\mathrm{C}$
3. The line passes through the y-axis, with coordinates in the x-z plane:${\mathrm{y}}_2=1\mathrm{m};{\mathrm{x}}_1=1\mathrm{m};{\mathrm{x}}_2=4\mathrm{m};{\mathrm{z}}_1=1\mathrm{m};{\mathrm{z}}_2=3\mathrm{m}$

Using the concept of the Gauss flux theorem, we can get the net flux through all the planes. This will help determine getting an expression for the electric charge, solving it will give the value of b in the electric field expression.

Formula:

The electric flux passing the surface,

$\mathrm{蠒}=鈭�\stackrel{鈫�}{\mathrm{E}}\stackrel{鈫�}{\mathrm{dA}}=\frac{\mathrm{q}}{{\mathrm{蔚}}_0}$ (1)

The net flux through the two faces parallel to the y-z plane is given using equation (1) such that,

$$\begin{gathered} {\mathrm{蠒}}_{\mathrm{yz}}=鈭�\left[{\mathrm{E}}_{\mathrm{x}2}-{\mathrm{E}}_{\mathrm{x}1}\right]\mathrm{dy}.\mathrm{dz} \\ ={鈭�}_0^1\mathrm{dy}{鈭�}_1^3\left[10+2\left(4\right)-10-2\left(1\right)\right]\mathrm{N}/\mathrm{C} \\ =\left(6\mathrm{N}/\mathrm{C}\right){鈭�}_0^1\mathrm{dy}{鈭�}_1^3\mathrm{dz} \\ =\left(6\mathrm{N}/\mathrm{C}\right)\left(1\mathrm{m}\right)\left(2\mathrm{m}\right) \\ =12\mathrm{N}路{\mathrm{m}}^2/\mathrm{C} \end{gathered}$$

Similarly, the net flux through the two faces parallel to the x-z plane is given using equation (i) as:

$$\begin{gathered} {\mathrm{蠒}}_{\mathrm{xz}}=鈭�\left[{\mathrm{E}}_{\mathrm{y}2}-{\mathrm{E}}_{\mathrm{y}1}\right]\mathrm{dx}.\mathrm{dz} \\ ={鈭�}_1^4\mathrm{dx}{鈭�}_1^3\mathrm{dz}\left[-3-\left(-3\right)\right]\mathrm{N}/\mathrm{C} \\ =0\mathrm{N}路{\mathrm{m}}^2/\mathrm{C} \end{gathered}$$

Now, the net flux through the two faces parallel to the x-y plane is given using equation (i) as:

$$\begin{gathered} {\mathrm{蠒}}_{\mathrm{xy}}=鈭�\left[{\mathrm{E}}_{\mathrm{z}2}-{\mathrm{E}}_{\mathrm{z}1}\right]\mathrm{dx}.\mathrm{dy} \\ ={鈭�}_1^4\mathrm{dx}{鈭�}_0^1\mathrm{dy}\left[3\mathrm{b}-\mathrm{b}\right]\mathrm{N}/\mathrm{C} \\ =\left(2\mathrm{b}\mathrm{N}/\mathrm{C}\right)\left(3\mathrm{m}\right)\left(1\mathrm{m}\right) \\ =6\mathrm{b}\mathrm{N}路{\mathrm{m}}^2/\mathrm{C} \end{gathered}$$

Now, the expression of the net charge due to the net flux can be given using equation (1) such that,

$$\begin{gathered} {\mathrm{q}}_{\mathrm{enc}}={\mathrm{蔚}}_0\mathrm{蠒} \\ {\mathrm{q}}_{\mathrm{enc}}={\mathrm{蔚}}_0\left({\mathrm{蠒}}_{\mathrm{xy}}+{\mathrm{蠒}}_{\mathrm{xz}}+{\mathrm{蠒}}_{\mathrm{yz}}\right) \\ {\mathrm{q}}_{\mathrm{enc}}={\mathrm{蔚}}_0\left(12\mathrm{N}路{\mathrm{m}}^2/\mathrm{C}+0\mathrm{N}路{\mathrm{m}}^2/\mathrm{C}+6\mathrm{b}\mathrm{N}路{\mathrm{m}}^2/\mathrm{C}\right) \\ 24.0{\mathrm{蔚}}_0\mathrm{C}=\left(12{\mathrm{蔚}}_0+6\mathrm{b}\right)\mathrm{N}路{\mathrm{m}}^2/\mathrm{C} \\ \mathrm{b}=12\mathrm{N}路{\mathrm{m}}^2/\mathrm{C} \end{gathered}$$

Hence, the required value is $12\mathrm{N}路{\mathrm{m}}^2/\mathrm{C}$.