91Ó°ÊÓ

The electric flux on the five faces of cube is${\mathrm{ϕ}}_5=564\text{V}·\text{m}$

The first charge inside the cube is$Q_1=9\text{nC}$

The second charge inside the cube is$Q_2=-6\text{nC}$

The third charge inside the cube is $Q_3=4\text{nC}$

The charges that are lying inside the cube can only contribute in the flux of the cube and those lying outside the cube do not contribute in the total flux.

Apply the Gauss’s law to find the total amount of electric flux of cube:

$\mathrm{Ï•}=\frac{Q_1+Q_2+Q_3}{{\mathrm{Î\mu }}_0}$

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.

$\begin{array}{rcl}\mathrm{ϕ}& =& \frac{9\text{nC}+\left(-6\text{nC}\right)+4\text{nC}}{8.854×{10}^{-12}{\text{C}}^2/\text{N}·{\text{m}}^2}\\ \mathrm{ϕ}& =& \frac{\left(7\text{nC}\right)\left(\frac{{10}^{-9}\text{C}}{1\text{nC}}\right)}{8.854×{10}^{-12}{\text{C}}^2/\text{N}·{\text{m}}^2}\\ \mathrm{ϕ}& =& 791\text{V}·\text{m}\\ & & \end{array}$

The electric flux through sixth face of cube is given as:

${\mathrm{Ï•}}_6=\mathrm{Ï•}-{\mathrm{Ï•}}_5$

Substitute all the values in the above equation.

$$\begin{gathered} {\mathrm{ϕ}}_6=791\text{V}·\text{m}-564\text{V}·\text{m} \\ {\mathrm{ϕ}}_6=227\text{V}·\text{m} \end{gathered}$$

Therefore, the electric flux through sixth face of cube is

$227V.m$ .