Chapter 24, Problem 24.32P

Assume the magnitude of the electric field on each face of the cube of edge L = 1.00 m in Figure P23.32 is uniform and the directions of the fields on each face are as indicated. Find (a) the net electric flux through the cube and (b) the net charge inside the cube. (c) Could the net charge he a single point charge?

Figure P23.32

To determine

The electric flux through the cube.

Answer to Problem 24.32P

The electric flux through the cube is $15.0\text{N}\cdot {\text{m}}^2/\text{C}$.

Explanation of Solution

The length of the edge of the cube is $1.00\text{m}$, the electric field on the each face of the cube is $25.0\text{N}/\text{C}$, $15.0\text{N}/\text{C}$, $35.0\text{N}/\text{C}$, $20.0\text{N}/\text{C}$ and $20.0\text{N}/\text{C}$.

The area vector is always out of face. The electric field going into the face of the cube makes $180°$ with the area vector and electric field going out of the face of the cube makes $0°$ with the area vector.

Write the expression for total flux through the cube

    ${\varphi }_{\text{E}}={\displaystyle \sum EA\cos \theta }$

Here, ${\varphi }_{\text{E}}$ is the net flux, $A$ is the area of the cylinder, $E$ is the electric field and $\theta$ is the angle between the electric field lines and the cube.

Write the area of the cube

    $A=L^2$

Here, $L$ is the length of the edge of the cube.

Substitute $L^2$ for $A$ in the above equation.

    $\begin{array}{c}{\varphi }_{\text{E}}={\displaystyle \sum E{L}^2\cos \theta }\\ =\left[\begin{array}{l}{E}_1{L}^2\cos \left(180°\right)+E_2{L}^2\cos \left(180°\right)+E_3{L}^2\cos \left(0°\right)+E_4{L}^2\cos \left(0°\right)\\ +E_5{L}^2\cos \left(0°\right)+E_6{L}^2\cos \left(0°\right)\end{array}\right]\end{array}$

Here, $E_1$, $E_2$, $E_3$, $E_4$, $E_5$ and $E_6$ are the flux through the face 1, 2, 3, 4, 5 and 6 of the cube respectively.

Conclusion:

Substitute $1.00\text{m}$ for $L$, $25.0\text{N}/\text{C}$ for $E_1$, $35.0\text{N}/\text{C}$ for $E_2$, $15.0\text{N}/\text{C}$ for $E_3$, $20.0\text{N}/\text{C}$ for $E_4$, $20.0\text{N}/\text{C}$ for $E_5$ and $20.0\text{N}/\text{C}$ for $E_6$ in above equation to find ${\varphi }_{\text{E}}$.

    $\begin{array}{c}{\varphi }_{\text{E}}=\left[\begin{array}{l}\left(25.0\text{N}/\text{C}\right){\left(1.00\text{m}\right)}^2\cos \left(180°\right)+\left(35.0\text{N}/\text{C}\right){\left(1.00\text{m}\right)}^2\cos \left(180°\right)\\ +\left(15.0\text{N}/\text{C}\right){\left(1.00\text{m}\right)}^2\cos \left(0°\right)+\left(20.0\text{N}/\text{C}\right){\left(1.00\text{m}\right)}^2\cos \left(0°\right)\\ +\left(20.0\text{N}/\text{C}\right){\left(1.00\text{m}\right)}^2\cos \left(0°\right)+\left(20.0\text{N}/\text{C}\right){\left(1.00\text{m}\right)}^2\cos \left(0°\right)\end{array}\right]\\ =\{{\left(1.00\text{m}\right)}^2\left[\begin{array}{l}\left(-25.0\text{N}/\text{C}\right)+\left(-35.0\text{N}/\text{C}\right)+\left(15.0\text{N}/\text{C}\right)\\ +\left(20.0\text{N}/\text{C}\right)+\left(20.0\text{N}/\text{C}\right)+\left(20.0\text{N}/\text{C}\right)\end{array}\right]\\ =15.0\text{N}\cdot {\text{m}}^2/\text{C}\end{array}$

Therefore, the electric flux through the cube is $15.0\text{N}\cdot {\text{m}}^2/\text{C}$.

To determine

The net charge inside the cube.

Answer to Problem 24.32P

The net charge inside the cube is $1.33\times {10}^{-10}\text{C}$.

Explanation of Solution

The length of the edge of the cube is $1.00\text{m}$, the electric field on the each face of the cube is $25.0\text{N}/\text{C}$, $15.0\text{N}/\text{C}$, $35.0\text{N}/\text{C}$, $20.0\text{N}/\text{C}$ and $20.0\text{N}/\text{C}$.

Write the expression for flux through the cube by Gauss law

    $\begin{array}{c}{\varphi }_{\text{E}}={\displaystyle \oint \overrightarrow{\text{E}}\cdot \text{d}\overrightarrow{\text{A}}}\\ =\frac{q_{\text{closed}}}{{\epsilon }_0}\\ q_{\text{closed}}={\varphi }_{\text{E}}{\epsilon }_0\end{array}$

Here, ${\epsilon }_0$ is the permittivity of free space and $q_{\text{closed}}$ is the charge enclosed by the cube.

Conclusion:

Substitute $15.0\text{N}\cdot {\text{m}}^2/\text{C}$ for ${\varphi }_{\text{E}}$ and $8.85\times {10}^{-12}\text{F}/\text{m}$ for ${\epsilon }_0$ in above equation to find $q_{\text{closed}}$.

    $\begin{array}{c}{q}_{\text{closed}}=\left(15.0\text{N}\cdot {\text{m}}^2/\text{C}\right)\left(8.85\times {10}^{-12}\text{F}/\text{m}\right)\\ =1.33\times {10}^{-10}\text{C}\end{array}$

Therefore, the net charge inside the cube is $1.33\times {10}^{-10}\text{C}$.

To determine

The number of charges inside the cube.

Answer to Problem 24.32P

The net charge is not a single point charge.

Explanation of Solution

The electric charge is the physical property of the matter that causes it to experience a force when placed in an electromagnetic field.

The net charge is not a single point charge as it does not produce a uniform field on the face of cubes. The equipotential surface of the single point charge is spherical. So the net charge is not a single point charge.

Conclusion:

Therefore, the net charge is not a single point charge.