91Ó°ÊÓ

The size of right and left face is $A=3cm×2cm$

The magnitude of electric field on the left face of box isrole="math" localid="1668579321256" $E_I=400V/m$ .

The magnitude of electric field on the right face of box is $E_r=1000V/m$.

The electric flux along the direction of electric field for box is at right and left face but the electric flux for other faces is zero because remaining all the faces are normal to the direction of electric field.

The electric flux at left face is given as:

${\mathrm{Ï•}}_l=-E_{l}A$

Substitute all the values in the above equation.

$$\begin{gathered} {\mathrm{ϕ}}_l=-\left(400\text{V}/\text{m}\right)\left(3\text{cm}×2\text{cm}\right)\left(\frac{1{\text{m}}^2}{{10}^4{\text{cm}}^2}\right) \\ {\mathrm{ϕ}}_l=-0.24\text{V}·\text{m} \end{gathered}$$

The electric flux at right face is given as:

${\mathrm{Ï•}}_r=E_{r}A$

Substitute all the values in the above equation.

$$\begin{gathered} {\mathrm{ϕ}}_r=\left(1000\text{V}/\text{m}\right)\left(3\text{cm}×2\text{cm}\right)\left(\frac{1{\text{m}}^2}{{10}^4{\text{cm}}^2}\right) \\ {\mathrm{ϕ}}_r=0.60\text{V}·\text{m} \end{gathered}$$

The electric flux from other faces of box is zero because all those faces are normal to electric field.

The total flux for the box is given as:

$\mathrm{Ï•}={\mathrm{Ï•}}_l+{\mathrm{Ï•}}_r$

Substitute all the values in the above equation.

$$\begin{gathered} \mathrm{ϕ}=-0.24\text{V}·\text{m}+0.60\text{V}·\text{m} \\ \mathrm{ϕ}=0.36\text{V}·\text{m} \end{gathered}$$

The amount of charge inside box is given as:

$\mathrm{Ï•}=\frac{q}{{\mathrm{Î\mu }}_0}$

Here, ${\mathrm{Î\mu }}_0$ is the permittivity of box and its value is $8.854×{10}^{-12}{\text{C}}^2/\text{N}·{\text{m}}^2$.

Substitute all the values in the above equation.

$\begin{array}{rcl}0.36\text{V}·\text{m}& =& \frac{q}{8.854×{10}^{-12}{\text{C}}^2/\text{N}·{\text{m}}^2}\\ q& ≈& 3.2×{10}^{-12}\text{C}\\ & & \end{array}$

Therefore, the electric flux on left face of box is$-0.24V.m$ , right face is $0.60V.m$, total flux through box is $0.36V.m$ and the amount of charge inside the box is $3.2×{10}^{-12}\text{C}$.