Question

Figure 21.62 shows a box on who surfaces the electric field is measured to be horizontal and to the right. On the left face (3 cm by 2 cm) the magnitude of electric field is 400 V/m and on the right face the magnitude of electric field is 1000 V/m. On the other faces only the direction is known (horizontal). Calculate the electric flux on every face of the box, the total flux and total amount of charge that is inside the box.

Short answer

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

Step-by-step solution

Identification of given data

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

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 $E_r=1000V/m$.

Conceptual Explanation

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.

Determination of electric flux left, right and total amount of flux

The electric flux at left face is given as:

${\varphi }_l=-E_{l}A$

Substitute all the values in the above equation.

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

The electric flux at right face is given as:

${\varphi }_r=E_{r}A$

Substitute all the values in the above equation.

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

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:

$\varphi ={\varphi }_l+{\varphi }_r$

Substitute all the values in the above equation.

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

Determination of amount of charge inside the box

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.

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

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 $3.2\times {10}^{-12}\text{C}$.