91Ó°ÊÓ

Given :

Charge : $10nC$

Dimensions of the cube : $2.0m×2.0m×2.0m$

Theory used :

The quantity of electric field that passes through a closed surface is referred to as the electric flux. The electric flux through a surface is proportional to the charge inside the surface, according to Gauss's law, which is given by :

${\mathrm{Φ}}_e=âˆ\circledR \stackrel{→}{E}·d\stackrel{→}{A}=\frac{Q_{in}}{{\mathrm{Î\mu }}_0}$ (1)

The electric flow is determined by the charge inside the closed surface, as indicated. Any flux owing to charges outside the closed surface is zero, thus we use the charges inside the cube, which are positive charges of $10nC$, to calculate the flux. The inert charge is :

role="math" localid="1649704858023" $$\begin{gathered} Q_{in}=(10nC)\frac{1×{10}^{-9}C}{nC} \\ =10×{10}^{-9}C \end{gathered}$$

To go within the closed surface, we plug the values for $Q_{in}\mathrm{and}{\mathrm{Î\mu }}_0$into equation (1)

$$\begin{gathered} {\mathrm{Φ}}_e=\frac{Q_{in}}{{\mathrm{Î\mu }}_0} \\ =\frac{10×{10}^{-9}C}{8.85×{10}^{-12}{C}^2/N{m}^2} \\ =\underline{1130N{m}^2/C} \end{gathered}$$

The cube has six faces, each of which has the same surface area. Because the electric flux inside the cube is determined by the charges, all sides have the same flux, so we can calculate the flux through the top face by dividing the total flux by six.

$$\begin{gathered} {\mathrm{Φ}}_{top}=\frac{{\mathrm{Φ}}_e}{6} \\ =\frac{1130N{m}^2/C}{6} \\ =0.19×{10}^{3}N{m}^2/C \\ =\boxed{\mathbf{0}\mathbf{.}\mathbf{19}\mathbf{}\mathit{k}\mathbf{}\mathit{N}\mathbf{·}{\mathit{m}}^{\mathbf{2}}\mathbf{/}\mathit{C}\mathbf{}} \end{gathered}$$