Question

A cube has sides of length$\mathbf{L}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{300}\mathbf{mm}$. One corner is at the origin (Fig. E22.6). The nonuniform electric field is given by$\overrightarrow{\mathbf{E}}\mathbf{=}\left(-5.00N/C\times m\right)\mathbf{x}\mathbf{+}\left(3.00N/C\times m\right)\mathbf{zk}$.(a) Find the electric flux through each of the six cube faces${\mathbf{S}}_{\mathbf{1}}\mathbf{,}{\mathbf{S}}_{\mathbf{2}}\mathbf{,}{\mathbf{S}}_{\mathbf{3}}\mathbf{,}{\mathbf{S}}_{\mathbf{4}}\mathbf{,}{\mathbf{S}}_{\mathbf{5}}$and${\mathbf{S}}_{\mathbf{6}}$. (b) Find the total electric charge inside the cube.

Short answer

(a)

The electric flux through the first side of the cube face surface is$0{\mathrm{Nm}}^2/\mathrm{C}$.
The electric flux through the second side of the cube face surface is$0.081{\mathrm{Nm}}^2/\mathrm{C}$
The electric flux through the third side of the cube face surface is$0{\mathrm{Nm}}^2/\mathrm{C}$
The electric flux through the fourth side of the cube face surface is$0{\mathrm{Nm}}^2/\mathrm{C}$
The electric flux through the fifth side of the cube face surface is$-0.135{\mathrm{Nm}}^2/\mathrm{C}$
The electric flux through the sixth side of the cube face surface is$0{\mathrm{Nm}}^2/\mathrm{C}$

(b)

The total electric charge inside the cube is$-4.781\times {10}^{-13}\mathrm{C}$.

Step-by-step solution

Define free body diagram and given data

The concept of electric flux, its calculation, and the analogy between the flux of an electric field and that of water. Let us imagine the flow of water with a velocity$\mathbf{v}$in a pipe in a fixed direction, say to the right. If we take the cross-sectional plane of the pipe and consider a small unit area given by ds from that plane, the volumetric flow of the liquid crossing that plane normal to the flow can be given as $\mathrm{v}\mathrm{ds}$.

Simplify the electrical flux

As equation$\left(22.8\right)$mentions, the electric flux through a spherical surface inside the inner surface of the sphere is given by:

${\mathrm{\varphi }}_{\mathrm{E}}=\frac{{\mathrm{Q}}_{\mathrm{enclosed}}}{{\mathrm{\epsilon }}_0}$

As equation $\left(22.5\right)$mentions, the electric flux through a surface is given by

localid="1665115999561" $$\begin{gathered} {\mathrm{\varphi }}_{\mathrm{E}}=\int \mathrm{EAcos}\left(\mathrm{f}\right) \\ =\int \overrightarrow{\mathrm{E}}\times \mathrm{d}\overrightarrow{\mathrm{A}} \end{gathered}$$

Solve for the first side of the cube face${\mathrm{S}}_1$:

$$\begin{gathered} {\mathrm{\varphi }}_{{\mathrm{S}}_1}=\overrightarrow{\mathrm{E}}\times {\mathrm{n}}_{{\mathrm{S}}_1}\mathrm{A} \\ =\overrightarrow{\mathrm{E}}\times \hat{\varnothing }\mathrm{A} \\ =0 \end{gathered}$$

So, the electric flux through the first side of the cube face surface is$0{\mathrm{Nm}}^2/\mathrm{C}$.

Solve for the second side of the cube face$S_2$:

localid="1665116435683" $$\begin{gathered} {\mathrm{\varphi }}_{{\mathrm{S}}_2}=\mathrm{E}\times {\hat{\mathrm{n}}}_{{\mathrm{S}}_2}\mathrm{A} \\ =\overrightarrow{\mathrm{E}}\times \hat{\mathrm{k}}\mathrm{A} \\ =3\mathrm{N}/\mathrm{C}\times \mathrm{m}\times \mathrm{z}\times {\left(0.3\mathrm{m}\right)}^2 \\ =3\mathrm{N}/\mathrm{C}\times \mathrm{m}\times 0.3\mathrm{m}\times {\left(0.3\mathrm{m}\right)}^2 \\ =3\mathrm{N}/\mathrm{C}\times \mathrm{m}\times {\left(0.3\mathrm{m}\right)}^3 \\ =0.081\mathrm{N}\times {\mathrm{m}}^2/\mathrm{C} \end{gathered}$$

So, the electric flux through the second side of the cube face surface is$0.081{\mathrm{Nm}}^2/\mathrm{C}$.
Solve for the third side of the cube face${\mathrm{S}}_3$:

$$\begin{gathered} {\mathrm{\varphi }}_{{\mathrm{S}}_3}=\overrightarrow{\mathrm{E}}\times {\hat{\mathrm{n}}}_3\mathrm{A} \\ =\overrightarrow{\mathrm{E}}\times \hat{\varnothing }\mathrm{A} \\ =0\mathrm{N}\times {\mathrm{m}}^2/\mathrm{C} \end{gathered}$$

So, the electric flux through the third side of the cube face surface is$0{\mathrm{Nm}}^2/\mathrm{C}$

Solve for the fourth side of the cube face$S_4$:

$$\begin{gathered} {\mathrm{\varphi }}_{{\mathrm{S}}_4}=\overrightarrow{\mathrm{E}}\times {\hat{\mathrm{n}}}_{{\mathrm{S}}_4}\mathrm{A} \\ =\overrightarrow{\mathrm{E}}\times \hat{\mathrm{k}}\mathrm{A} \\ =-3\mathrm{N}/\mathrm{C}\times \mathrm{m}\times \mathrm{z}\times {\left(0.3\mathrm{m}\right)}^2 \\ =-3\mathrm{N}/\mathrm{C}\times \mathrm{m}\times 0\mathrm{m}\times {\left(0.3\mathrm{m}\right)}^2 \\ =0\mathrm{N}\times {\mathrm{m}}^2/\mathrm{C} \end{gathered}$$

So, the electric flux through the fourth side of the cube face surface is$0{\mathrm{Nm}}^2/\mathrm{C}$.
Solve for the fifth side of the cube face${\mathrm{S}}_5$:

$$\begin{gathered} {\mathrm{\varphi }}_{{\mathrm{S}}_5}=\overrightarrow{\mathrm{E}}\times {\hat{\mathrm{n}}}_{{\mathrm{S}}_5}\mathrm{A} \\ =\overrightarrow{\mathrm{E}}\times \hat{\mathrm{l}}\mathrm{A} \\ =-5\mathrm{N}/\mathrm{C}\times \mathrm{m}\times \mathrm{x}\times {\left(0.3\mathrm{m}\right)}^2 \\ =-5\mathrm{N}/\mathrm{C}\times \mathrm{m}\times 0.3\mathrm{m}\times {\left(0.3\mathrm{m}\right)}^2 \\ =-5\mathrm{N}/\mathrm{C}\times \mathrm{m}\times {\left(0.3\mathrm{m}\right)}^3 \\ =-0.135\mathrm{N}\times {\mathrm{m}}^2/\mathrm{C} \end{gathered}$$

So, the electric flux through the fifth side of the cube face surface is $-0.135{\mathrm{Nm}}^2/\mathrm{C}$.

Solve for the sixth side of the cube face${\mathrm{S}}_6$:

$$\begin{gathered} {\mathrm{\varphi }}_{{\mathrm{S}}_6}=\overrightarrow{\mathrm{E}}\times {\hat{\mathrm{S}}}_6\mathrm{A} \\ =-\overrightarrow{\mathrm{E}}\times \hat{\mathrm{\iota }}\mathrm{A} \\ =-5\mathrm{N}/\mathrm{C}\times \mathrm{m}\times \mathrm{x}\times {\left(0.3\mathrm{m}\right)}^2 \\ =-5\mathrm{N}/\mathrm{C}\times \mathrm{m}\times 0\mathrm{m}\times {\left(0.3\mathrm{m}\right)}^2 \\ =0\mathrm{N}\times {\mathrm{m}}^2/\mathrm{C} \end{gathered}$$

So, the electric flux through the sixth side of the cube face surface is$0{\mathrm{Nm}}^2/\mathrm{C}$.

Solve total electric charge

Solve for the total electric charge inside the cube:
We will take the data (electric flux through each side of the cube) from part (a). In order to evaluate the total electric charge density inside the cube, we use the following relation:

$$\begin{gathered} {\mathrm{\Phi }}_{\mathrm{total}}=\sum {\mathrm{\Phi }}_{\mathrm{S}} \\ ={\mathrm{\Phi }}_{{\mathrm{S}}_1}+{\mathrm{\Phi }}_{{\mathrm{S}}_2}+{\mathrm{\Phi }}_{{\mathrm{S}}_3}+{\mathrm{\Phi }}_{{\mathrm{S}}_4}+{\mathrm{\Phi }}_{{\mathrm{S}}_5}+{\mathrm{\Phi }}_{{\mathrm{S}}_6} \\ =\left(0+0.081+0+0+-0.135+0\right)N{m}^2/C \\ =-0.054N{\mathrm{m}}^2/C \end{gathered}$$

As equation $\left(22.8\right)$mentions, the electric flux through a spherical surface inside the inner surface of the sphere is given by:

$$\begin{gathered} {\mathrm{\Phi }}_{\mathrm{total}}==\sum {\mathrm{\Phi }}_{\mathrm{S}} \\ ={\mathrm{\Phi }}_{{\mathrm{S}}_1}+{\mathrm{\Phi }}_{{\mathrm{S}}_2}+{\mathrm{\Phi }}_{{\mathrm{S}}_3}+{\mathrm{\Phi }}_{{\mathrm{S}}_4}+{\mathrm{\Phi }}_{{\mathrm{S}}_5}+{\mathrm{\Phi }}_{{\mathrm{S}}_6} \\ =\left(0+0.081+0+0+-0.135+0\right)N{\mathrm{m}}^2/C \\ =-0.054N{\mathrm{m}}^2/C \end{gathered}$$

So, the total electric charge inside the cube is$-4.781\times {10}^{-13}\mathrm{C}$.