The center of a thin paper cube in outer space is located at the origin. Each edge is 10 cm long. The only other objects in the neighborhood are some small charged particles whose charges and position at this instant are following: +9 nC at (4,2,-3) cm, -6 nC at (1,-3,2) cm, +7 nC at (15,0,4) cm, +4 nC at (-1,3,2) cm and -3 nC at (-2,12,-3) cm. The electric flux on the five of the six faces of the cube totals 564 V m. What is the flux on the other face?

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

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:

$\varphi =\frac{Q_1+Q_2+Q_3}{{\epsilon }_0}$

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

Substitute all the values in the above equation.

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

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

${\varphi }_6=\varphi -{\varphi }_5$

Substitute all the values in the above equation.

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

Therefore, the electric flux through sixth face of cube is

$227V.m$ .