Chapter 24, Problem 32P

(a)

To determine

The net electric flux through the cube.

Answer to Problem 32P

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

Explanation of Solution

Below figure shows the electric field magnitude and its direction in all face’s of the cube.

Figure (1)

From Figure (1), it is shown that, the electric fields are perpendicular to the faces of the cube. Therefore, use Gauss law for net flux through the closed Gaussian surface.

Write the expression for net flux through the closed Gaussian surface.

    ${\varphi }_{\text{E}}={\displaystyle \oint \overrightarrow{E}\cdot d\overrightarrow{A}}$                                                                                                             (I)

Here, $\overrightarrow{E}$ is the electric field, ${\varphi }_{\text{E}}$ is the net flux through closed Gaussian surface and $\overrightarrow{A}$ is the surface area.

Write the expression for net electric flux through cube.

    ${\varphi }_{\text{E}}={\varphi }_{\text{E}\left(\text{top}\right)}+{\varphi }_{\text{E}\left(\text{bottom}\right)}+{\varphi }_{\text{E}\left(\text{right}\right)}+{\varphi }_{\text{E}\left(\text{left}\right)}+{\varphi }_{\text{E}\left(\text{into}\right)}+{\varphi }_{\text{E}\left(\text{out}\right)}$                                              (II)

Conclusion:

Substitute $25.0\text{N}/\text{C}$ for $E$, and $L^2$ for $A$ in equation (I) to calculate ${\varphi }_{\text{E}\left(\text{top}\right)}$.

    $\begin{array}{c}{\varphi }_{\text{E}\left(\text{top}\right)}=EA\cos \left(180°\right)\\ =-\left(EA\right)\\ =-\left(25.0\text{N}/\text{C}\right)\left(1.00{\text{m}}^2\right)\\ =-25.0\text{N}\cdot {\text{m}}^2/\text{C}\end{array}$

Substitute $15.0\text{N}/\text{C}$ for $E$, and $L^2$ for $A$ in equation (I) to calculate ${\varphi }_{\text{E}\left(\text{bottom}\right)}$

    $\begin{array}{c}{\varphi }_{\text{E}\left(\text{bottom}\right)}=EA\cos \left(0°\right)\\ =EA\\ =\left(15.0\text{N}/\text{C}\right)\left(1.00{\text{m}}^2\right)\\ =15.0\text{N}\cdot {\text{m}}^2/\text{C}\end{array}$

Substitute $35.0\text{N}/\text{C}$ for $E$, and $L^2$ for $A$ in equation (I) to calculate ${\varphi }_{\text{E}\left(\text{right}\right)}$.

    $\begin{array}{c}{\varphi }_{\text{E}\left(\text{right}\right)}=EA\cos \left(180°\right)\\ =-\left(EA\right)\\ =-\left(35.0\text{N}/\text{C}\right)\left(1.00{\text{m}}^2\right)\\ =-35.0\text{N}\cdot {\text{m}}^2/\text{C}\end{array}$

Substitute $20.0\text{N}/\text{C}$ for $E$, and $L^2$ for $A$ in equation (I) to calculate ${\varphi }_{\text{E}\left(\text{left}\right)}$.

    $\begin{array}{c}{\varphi }_{\text{E}\left(\text{left}\right)}=EA\cos \left(0°\right)\\ =EA\\ =\left(20.0\text{N}/\text{C}\right)\left(1.00{\text{m}}^2\right)\\ =20.0\text{N}\cdot {\text{m}}^2/\text{C}\end{array}$

Substitute $20.0\text{N}/\text{C}$ for $E$, and $L^2$ for $A$ in equation (I) to calculate ${\varphi }_{\text{E}\left(\text{into}\right)}$.

    $\begin{array}{c}{\varphi }_{\text{E}\left(\text{into}\right)}=EA\cos \left(0°\right)\\ =EA\\ =\left(20.0\text{N}/\text{C}\right)\left(1.00{\text{m}}^2\right)\\ =20.0\text{N}\cdot {\text{m}}^2/\text{C}\end{array}$

Substitute $20.0\text{N}/\text{C}$ for $E$, and $L^2$ for $A$ in equation (I) to calculate ${\varphi }_{\text{E}\left(\text{out}\right)}$.

    $\begin{array}{c}{\varphi }_{\text{E}\left(\text{out}\right)}=EA\cos \left(0°\right)\\ =EA\\ =\left(20.0\text{N}/\text{C}\right)\left(1.00{\text{m}}^2\right)\\ =20.0\text{N}\cdot {\text{m}}^2/\text{C}\end{array}$

Substitute $-25.0\text{N}\cdot {\text{m}}^2/\text{C}$ for ${\varphi }_{\text{E}\left(\text{top}\right)}$, $15.0\text{N}\cdot {\text{m}}^2/\text{C}$ for ${\varphi }_{\text{E}\left(\text{bottom}\right)}$, $-35.0\text{N}\cdot {\text{m}}^2/\text{C}$ for ${\varphi }_{\text{E}\left(\text{right}\right)}$, $20.0\text{N}\cdot {\text{m}}^2/\text{C}$ for ${\varphi }_{\text{E}\left(\text{left}\right)}$, ${\varphi }_{\text{E}\left(\text{into}\right)}$ and ${\varphi }_{\text{E}\left(\text{out}\right)}$ in equation (II) to calculate ${\varphi }_{\text{E}}$.

    $\begin{array}{c}{\varphi }_{\text{E}}=\left(-25.0+15.0-35.0+20.0+20.0+20.0\right)\text{N}\cdot {\text{m}}^2/\text{C}\\ =15.0\text{N}\cdot {\text{m}}^2/\text{C}\end{array}$

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

(b)

To determine

The net charge inside the cube.

Answer to Problem 32P

The charge enclosed within the Gaussian surface (cube) is $0.133\times {10}^{-9}\text{C}$.

Explanation of Solution

Write the expression for the charge enclosed within the Gaussian surface (cube) is.

    $q={\epsilon }_0{\varphi }_{\text{E}}$                                                                                                                 (II)

Here, $q$ is the charge enclosed within the Gaussian surface (cube), ${\epsilon }_0$ is the permittivity of free space and ${\varphi }_{\text{E}}$ is the net electric flux through the cube.

Conclusion:

Substitute $8.85\times {10}^{-12}{\text{C}}^2/\text{N}\cdot {\text{m}}^2$ for ${\epsilon }_0$, $15.0\text{N}\cdot {\text{m}}^2/\text{C}$ for ${\varphi }_{\text{E}}$ in Equation (II) to calculate $q$.

    $\begin{array}{c}q=\left(8.85\times {10}^{-12}{\text{C}}^2/\text{N}\cdot {\text{m}}^2\right)\left(15.0\text{N}\cdot {\text{m}}^2/\text{C}\right)\\ =0.133\times {10}^{-9}\text{C}\end{array}$

Therefore, the charge enclosed within the Gaussian surface (cube) is $0.133\times {10}^{-9}\text{C}$.

(c)

To determine

Whether the net charge could be a point charge.

Answer to Problem 32P

No, the net charge can’t be a point charge.

Explanation of Solution

No, single positive charge’s magnitude is $q^{\prime }=1.602\times {10}^{-19}\text{C}$, which is not equal to the obtained net charge $q=0.133\times {10}^{-9}\text{C}$.

Conclusion:

Therefore, the net charge can’t be a point charge.