Question

The cube in Fig. 23-31 has edge length 1.40mand is oriented as shown in a region of uniform electric field. Find the electric flux through the right face if the electric field, in Newton per coulomb, is given by (a) $\mathbf{6}\mathbf{.}\mathbf{00}\hat{\mathbf{i}}$,(b) $\mathbf{-}\mathbf{2}\mathbf{.}\mathbf{00}\hat{\mathbf{j}}$, and $\mathbf{-}\mathbf{3}\mathbf{.}\mathbf{00}\hat{\mathbf{i}}\mathbf{+}\mathbf{4}\mathbf{.}\mathbf{00}\hat{\mathbf{k}}$(c). (d) What is the total flux through the cube for each field?

Short answer

1. The electric flux through the right face for the electric field is zero.
2. The electric flux through the right face for electric field is$3.92\mathrm{N}.{\mathrm{m}}^2/\mathrm{C}$.
3. The electric flux through the right face for the electric field is zero.
4. The total flux through the cube for each field is zero.

Step-by-step solution

The given data

1. The edge length of the cube,$\mathrm{a}=1.4\mathrm{m}$
2. Electric fields, $6.00\hat{\mathrm{i}}$,$-2.00\hat{\mathrm{j}}$, and$-3.00\hat{\mathrm{i}}+4.00\hat{\mathrm{k}}$

Understanding the concept of Gauss law-planar symmetry

Using the concept of the Gauss flux theorem, the flux value through each surface for the given electric field can be calculated. Again, the total flux within an enclosed volume for a uniform electric field is always zero.

Formula:

The electric flux through any enclosed surface, $\mathrm{\varphi }=\overrightarrow{\mathrm{E}}.\overrightarrow{\mathrm{dA}}$ (1)

a) Calculation of the electric flux

The area of the surface is given as:

$$\begin{gathered} \overrightarrow{\mathrm{A}}=\mathrm{A}\hat{\mathrm{j}} \\ ={\left(1.40\mathrm{m}\right)}^2\hat{\mathrm{j}} \end{gathered}$$

Thus, the value of the electric flux is given using equation (1) as:

role="math" localid="1657345803012" $$\begin{gathered} \mathrm{\varphi }=\left(\left(6.00\mathrm{N}/\mathrm{C}\right)\hat{\mathrm{i}}\right).\left({\left(1.40\mathrm{m}\right)}^2\hat{\mathrm{j}}\right) \\ =0\mathrm{N}.{\mathrm{m}}^2/\mathrm{C} \end{gathered}$$

Hence, the value of the electric flux is zero.

b) Calculation of the electric flux

The value of the electric flux is given using equation (1) as:

$$\begin{gathered} \mathrm{\varphi }=\left(\left(-2.00\mathrm{N}/\mathrm{C}\right)\hat{\mathrm{j}}\right).{\left(1.40\mathrm{m}\right)}^2\hat{\mathrm{j}} \\ =3.92\mathrm{N}.{\mathrm{m}}^2/\mathrm{C} \end{gathered}$$

Hence, the value of the electric flux is $3.92\mathrm{N}.{\mathrm{m}}^2/\mathrm{C}$.

c) Calculation of the electric flux

The value of the electric flux is given using equation (i) as:

$$\begin{gathered} \mathrm{\varphi }=\left[\left(-3.00\mathrm{N}/\mathrm{C}\right)\hat{\mathrm{i}}+\left(400\mathrm{N}/\mathrm{C}\right)\hat{\mathrm{k}}\right].{\left(1.40\mathrm{m}\right)}^2\hat{\mathrm{j}} \\ =0\mathrm{N}.{\mathrm{m}}^2/\mathrm{C} \end{gathered}$$

Hence, the value of the electric flux is zero.

Calculation of the total flux through the cube for each field

The total flux of a uniform field through a closed surface is always zero.