Question

The electric field has been measured to be vertically upward everywhere on the surface of a box 30 cm long, 4 cm high and 3 cm deep as shown in Figure 21.64. All over the bottom of the box ${\mathbf{E}}_{\mathbf{1}}\mathbf{=}\mathbf{1500}\mathbf{V}\mathbf{/}\mathbf{m}$ , all over the sides ${\mathbf{E}}_{\mathbf{2}}\mathbf{=}\mathbf{1000}\mathbf{V}\mathbf{/}\mathbf{m}$and all over the top ${\mathbf{E}}_{\mathbf{3}}\mathbf{=}\mathbf{600}\mathbf{V}\mathbf{/}\mathbf{m}$. What can you conclude about the contents of the box. Include a numerical result.

Short answer

We can conclude from the above electric field distribution that box contains the charge$-5\times {10}^{-11}\text{C}$ between inside surface of box.

Step-by-step solution

Identification of given data

The length of box is $l=20cm$

The depth of box is$d=3cm$ .

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

The magnitude of electric field on the bottom of box is$E_1=1500V/m$ .

The magnitude of electric field on the sides of box is $E_2=1000V/m$.

The magnitude of electric field on the top of box is $E_3=600V/m$.

Conceptual Explanation

The net electric flux from top and bottom 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.

Determination of charge enclosed inside the box

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

$\varphi =\left(E_3-E_1\right)ld$

Substitute all the values in the above equation.

$$\begin{gathered} \varphi =\left(600\text{V}/\text{m}-1500\text{V}/\text{m}\right)\left[\left(20\text{cm}\right)\left(\frac{1\text{m}}{{10}^2\text{cm}}\right)\right]\left[\left(3\text{cm}\right)\left(\frac{1\text{m}}{{10}^2\text{cm}}\right)\right] \\ \varphi =-5.4\text{V}·\text{m} \end{gathered}$$

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.

$\begin{array}{rcl}-5.4\text{V}·\text{m}& =& \frac{q}{8.854\times {10}^{-12}{\text{C}}^2/\text{N}·{\text{m}}^2}\\ q& =& -4.78\times {10}^{-11}\text{C}\\ q& \approx & -5\times {10}^{-11}\text{C}\\ & & \end{array}$

Therefore, the amount of charge inside the box is $-5\times {10}^{-11}\text{C}$.