The electric field has been measured to be horizontal and to the right everywhere on the closed box as shown in Figure 21.66. All over the left side of box ${\mathbf{E}}_{\mathbf{1}}\mathbf{=}\mathbf{100}\mathbf{V}\mathbf{/}\mathbf{m}$ and all over the right (slanting) side of box ${\mathbf{E}}_{\mathbf{2}}\mathbf{=}\mathbf{300}\mathbf{}\mathbf{V}\mathbf{/}\mathbf{m}$.On the top the average field is ${\mathbf{E}}_{\mathbf{3}}\mathbf{=}\mathbf{150}\mathbf{V}\mathbf{/}\mathbf{m}$ , on the front and back the average field is ${\mathbf{E}}_{\mathbf{4}}\mathbf{=}\mathbf{175}\mathbf{V}\mathbf{/}\mathbf{m}$and on the bottom the average field is ${\mathbf{E}}_{\mathbf{5}}\mathbf{=}\mathbf{220}\mathbf{V}\mathbf{/}\mathbf{m}$.How much charge is inside the box? Explain briefly.

The length of box is $l=18cm$

The width of box is $d=5cm$.

The height of box is $h=4cm$.

The length of slanting side of right face of box is$b=12cm$

The magnitude of electric field on the left of box isrole="math" localid="1668577036373" $E_1=100V/m$ .

The magnitude of electric field on the right side of box is $E_2=300V/m$.

The net electric flux from left and right sides of box is calculated, then Gauss law is applied for the box to find enclosed charge inside the box. The electric flux from all other sides is zero because all the remaining surfaces are normal to the electric field for those surfaces.

The net electric flux from top and bottom is given as:

$\varphi =\left(E_2-E_1\right)hd$

Substitute all the values in the above equation.$$\begin{gathered} \varphi =\left(300\text{V}/\text{m}-100\text{V}/\text{m}\right)\left[\left(4\text{cm}\right)\left(\frac{1\text{m}}{{10}^2\text{cm}}\right)\right]\left[\left(5\text{cm}\right)\left(\frac{1\text{m}}{{10}^2\text{cm}}\right)\right] \\ \varphi =0.4\text{V}·\text{m} \end{gathered}$$

Apply the Gauss’s law to find the amount of charge inside box is given as:

$\varphi =\frac{q}{{\epsilon }_0}$

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.

$\begin{array}{rcl}0.4\text{V}·\text{m}& =& \frac{q}{8.854\times {10}^{-12}{\text{C}}^2/\text{N}·{\text{m}}^2}\\ q& =& 3.541\times {10}^{-12}\text{C}\\ & & \end{array}$

Therefore, the amount of charge inside the box is $3.541\times {10}^{-12}\text{C}$.