91Ó°ÊÓ

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$-3.00\hat{\mathrm{i}}+4.00\hat{\mathrm{k}}$

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, $\mathrm{ϕ}=\stackrel{→}{\mathrm{E}}.\stackrel{→}{\mathrm{dA}}$ (1)

The area of the surface is given as:

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

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

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

Hence, the value of the electric flux is zero.

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

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

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

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

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

Hence, the value of the electric flux is zero.

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